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T-distribution, f-distribution.

9.4 Full Hypothesis Test Examples

Tests on means, example 9.8.

Jeffrey, as an eight-year old, established a mean time of 16.43 seconds for swimming the 25-yard freestyle, with a standard deviation of 0.8 seconds . His dad, Frank, thought that Jeffrey could swim the 25-yard freestyle faster using goggles. Frank bought Jeffrey a new pair of expensive goggles and timed Jeffrey for 15 25-yard freestyle swims . For the 15 swims, Jeffrey's mean time was 16 seconds. Frank thought that the goggles helped Jeffrey to swim faster than the 16.43 seconds. Conduct a hypothesis test using a preset α = 0.05. Assume that the swim times for the 25-yard freestyle are normal.

Set up the Hypothesis Test:

Since the problem is about a mean, this is a test of a single population mean .

H 0 : μ = 16.43   H a : μ < 16.43

For Jeffrey to swim faster, his time will be less than 16.43 seconds. The "<" tells you this is left-tailed.

Determine the distribution needed:

Random variable: X ¯ X ¯ = the mean time to swim the 25-yard freestyle.

Distribution for the test: X ¯ X ¯ is normal (population standard deviation is known: σ = 0.8)

X ¯ ~ N ( μ , σ X n ) X ¯ ~ N ( μ , σ X n ) Therefore, X ¯ ~ N ( 16.43 , 0.8 15 ) X ¯ ~ N ( 16.43 , 0.8 15 )

μ = 16.43 comes from H 0 and not the data. σ = 0.8, and n = 15.

Calculate the p -value using the normal distribution for a mean:

p -value = P ( x ¯ x ¯ < 16) = 0.0187 where the sample mean in the problem is given as 16.

p -value = 0.0187 (This is called the actual level of significance .) The p -value is the area to the left of the sample mean is given as 16.

μ = 16.43 comes from H 0 . Our assumption is μ = 16.43.

Interpretation of the p -value: If H 0 is true , there is a 0.0187 probability (1.87%)that Jeffrey's mean time to swim the 25-yard freestyle is 16 seconds or less. Because a 1.87% chance is small, the mean time of 16 seconds or less is unlikely to have happened randomly. It is a rare event.

Compare α and the p -value:

α = 0.05 p -value = 0.0187 α > p -value

Make a decision: Since α > α > p -value, reject H 0 .

This indicates that you reject the null hypothesis that the mean time to swim the 25-yard freestyle is at least 16.43 seconds.

Conclusion: At the 5% significance level, there is sufficient evidence that Jeffrey's mean time to swim the 25-yard freestyle is less than 16.43 seconds. Thus, based on the sample data, we conclude that Jeffrey swims faster using the new goggles.

The Type I and Type II errors for this problem are as follows: The Type I error is to conclude that Jeffrey swims the 25-yard freestyle, on average, in less than 16.43 seconds when, in fact, he actually swims the 25-yard freestyle, on average, in at least 16.43 seconds. (Reject the null hypothesis when the null hypothesis is true.)

The Type II error is that there is not evidence to conclude that Jeffrey swims the 25-yard freestyle, on average, in less than 16.43 seconds when, in fact, he actually does swim the 25-yard free-style, on average, in less than 16.43 seconds. (Do not reject the null hypothesis when the null hypothesis is false.)

The mean throwing distance of a football for Marco, a high school quarterback, is 40 yards, with a standard deviation of two yards. The team coach tells Marco to adjust his grip to get more distance. The coach records the distances for 20 throws. For the 20 throws, Marco’s mean distance was 45 yards. The coach thought the different grip helped Marco throw farther than 40 yards. Conduct a hypothesis test using a preset α = 0.05. Assume the throw distances for footballs are normal.

First, determine what type of test this is, set up the hypothesis test, find the p -value, sketch the graph, and state your conclusion.

Example 9.9

Jasmine has just begun her new job on the sales force of a very competitive company. In a sample of 16 sales calls it was found that she closed the contract for an average value of 108 dollars with a standard deviation of 12 dollars. Test at 5% significance that the population mean is at least 100 dollars against the alternative that it is less than 100 dollars. Company policy requires that new members of the sales force must exceed an average of $100 per contract during the trial employment period. Can we conclude that Jasmine has met this requirement at the significance level of 95%?

• H 0 : µ ≤ 100 H a : µ > 100 The null and alternative hypothesis are for the parameter µ because the number of dollars of the contracts is a continuous random variable. Also, this is a one-tailed test because the company has only an interested if the number of dollars per contact is below a particular number not "too high" a number. This can be thought of as making a claim that the requirement is being met and thus the claim is in the alternative hypothesis.
• Test statistic: t c = x ¯ − µ 0 s n = 108 − 100 ( 12 16 ) = 2.67 t c = x ¯ − µ 0 s n = 108 − 100 ( 12 16 ) = 2.67
• Critical value: t a = 1.753 t a = 1.753 with n-1 degrees of freedom= 15

The test statistic is a Student's t because the sample size is below 30; therefore, we cannot use the normal distribution. Comparing the calculated value of the test statistic and the critical value of t t ( t a ) ( t a ) at a 5% significance level, we see that the calculated value is in the tail of the distribution. Thus, we conclude that 108 dollars per contract is significantly larger than the hypothesized value of 100 and thus we cannot accept the null hypothesis. There is evidence that supports Jasmine's performance meets company standards.

It is believed that a stock price for a particular company will grow at a rate of $5 per week with a standard deviation of $1. An investor believes the stock won’t grow as quickly. The changes in stock price is recorded for ten weeks and are as follows: $4, $3, $2, $3, $1, $7, $2, $1, $1, $2. Perform a hypothesis test using a 5% level of significance. State the null and alternative hypotheses, state your conclusion, and identify the Type I errors.

Example 9.10

A manufacturer of salad dressings uses machines to dispense liquid ingredients into bottles that move along a filling line. The machine that dispenses salad dressings is working properly when 8 ounces are dispensed. Suppose that the average amount dispensed in a particular sample of 35 bottles is 7.91 ounces with a variance of 0.03 ounces squared, s 2 s 2 . Is there evidence that the machine should be stopped and production wait for repairs? The lost production from a shutdown is potentially so great that management feels that the level of significance in the analysis should be 99%.

Again we will follow the steps in our analysis of this problem.

STEP 1 : Set the Null and Alternative Hypothesis. The random variable is the quantity of fluid placed in the bottles. This is a continuous random variable and the parameter we are interested in is the mean. Our hypothesis therefore is about the mean. In this case we are concerned that the machine is not filling properly. From what we are told it does not matter if the machine is over-filling or under-filling, both seem to be an equally bad error. This tells us that this is a two-tailed test: if the machine is malfunctioning it will be shutdown regardless if it is from over-filling or under-filling. The null and alternative hypotheses are thus:

STEP 2 : Decide the level of significance and draw the graph showing the critical value.

This problem has already set the level of significance at 99%. The decision seems an appropriate one and shows the thought process when setting the significance level. Management wants to be very certain, as certain as probability will allow, that they are not shutting down a machine that is not in need of repair. To draw the distribution and the critical value, we need to know which distribution to use. Because this is a continuous random variable and we are interested in the mean, and the sample size is greater than 30, the appropriate distribution is the normal distribution and the relevant critical value is 2.575 from the normal table or the t-table at 0.005 column and infinite degrees of freedom. We draw the graph and mark these points.

STEP 3 : Calculate sample parameters and the test statistic. The sample parameters are provided, the sample mean is 7.91 and the sample variance is .03 and the sample size is 35. We need to note that the sample variance was provided not the sample standard deviation, which is what we need for the formula. Remembering that the standard deviation is simply the square root of the variance, we therefore know the sample standard deviation, s, is 0.173. With this information we calculate the test statistic as -3.07, and mark it on the graph.

STEP 4 : Compare test statistic and the critical values Now we compare the test statistic and the critical value by placing the test statistic on the graph. We see that the test statistic is in the tail, decidedly greater than the critical value of 2.575. We note that even the very small difference between the hypothesized value and the sample value is still a large number of standard deviations. The sample mean is only 0.08 ounces different from the required level of 8 ounces, but it is 3 plus standard deviations away and thus we cannot accept the null hypothesis.

STEP 5 : Reach a Conclusion

Three standard deviations of a test statistic will guarantee that the test will fail. The probability that anything is within three standard deviations is almost zero. Actually it is 0.0026 on the normal distribution, which is certainly almost zero in a practical sense. Our formal conclusion would be “ At a 99% level of significance we cannot accept the hypothesis that the sample mean came from a distribution with a mean of 8 ounces” Or less formally, and getting to the point, “At a 99% level of significance we conclude that the machine is under filling the bottles and is in need of repair”.

Try It 9.10

A company records the mean time of employees working in a day. The mean comes out to be 475 minutes, with a standard deviation of 45 minutes. A manager recorded times of 20 employees. The times of working were (frequencies are in parentheses) 460(3); 465(2); 470(3); 475(1); 480(6); 485(3); 490(2).

Conduct a hypothesis test using a 2.5% level of significance to determine if the mean time is more than 475 .

Hypothesis Test for Proportions

Just as there were confidence intervals for proportions, or more formally, the population parameter p of the binomial distribution, there is the ability to test hypotheses concerning p .

The population parameter for the binomial is p . The estimated value (point estimate) for p is p′ where p′ = x/n , x is the number of successes in the sample and n is the sample size.

When you perform a hypothesis test of a population proportion p , you take a simple random sample from the population. The conditions for a binomial distribution must be met, which are: there are a certain number n of independent trials meaning random sampling, the outcomes of any trial are binary, success or failure, and each trial has the same probability of a success p . The shape of the binomial distribution needs to be similar to the shape of the normal distribution. To ensure this, the quantities np′ and nq′ must both be greater than five ( np′ > 5 and nq′ > 5). In this case the binomial distribution of a sample (estimated) proportion can be approximated by the normal distribution with μ = np μ = np and σ = npq σ = npq . Remember that q = 1 – p q = 1 – p . There is no distribution that can correct for this small sample bias and thus if these conditions are not met we simply cannot test the hypothesis with the data available at that time. We met this condition when we first were estimating confidence intervals for p .

Again, we begin with the standardizing formula modified because this is the distribution of a binomial.

Substituting p 0 p 0 , the hypothesized value of p , we have:

This is the test statistic for testing hypothesized values of p , where the null and alternative hypotheses take one of the following forms:

The decision rule stated above applies here also: if the calculated value of Z c shows that the sample proportion is "too many" standard deviations from the hypothesized proportion, the null hypothesis cannot be accepted. The decision as to what is "too many" is pre-determined by the analyst depending on the level of significance required in the test.

Example 9.11

The mortgage department of a large bank is interested in the nature of loans of first-time borrowers. This information will be used to tailor their marketing strategy. They believe that 50% of first-time borrowers take out smaller loans than other borrowers. They perform a hypothesis test to determine if the percentage is the same or different from 50% . They sample 100 first-time borrowers and find 53 of these loans are smaller that the other borrowers. For the hypothesis test, they choose a 5% level of significance.

STEP 1 : Set the null and alternative hypothesis.

H 0 : p = 0.50   H a : p ≠ 0.50

The words "is the same or different from" tell you this is a two-tailed test. The Type I and Type II errors are as follows: The Type I error is to conclude that the proportion of borrowers is different from 50% when, in fact, the proportion is actually 50%. (Reject the null hypothesis when the null hypothesis is true). The Type II error is there is not enough evidence to conclude that the proportion of first time borrowers differs from 50% when, in fact, the proportion does differ from 50%. (You fail to reject the null hypothesis when the null hypothesis is false.)

STEP 2 : Decide the level of significance and draw the graph showing the critical value

The level of significance has been set by the problem at the 5% level. Because this is two-tailed test one-half of the alpha value will be in the upper tail and one-half in the lower tail as shown on the graph. The critical value for the normal distribution at the 95% level of confidence is 1.96. This can easily be found on the student’s t-table at the very bottom at infinite degrees of freedom remembering that at infinity the t-distribution is the normal distribution. Of course the value can also be found on the normal table but you have go looking for one-half of 95 (0.475) inside the body of the table and then read out to the sides and top for the number of standard deviations.

STEP 3 : Calculate the sample parameters and critical value of the test statistic.

The test statistic is a normal distribution, Z, for testing proportions and is:

For this case, the sample of 100 found 53 of these loans were smaller than those of other borrowers. The sample proportion, p′ = 53/100= 0.53 The test question, therefore, is : “Is 0.53 significantly different from .50?” Putting these values into the formula for the test statistic we find that 0.53 is only 0.60 standard deviations away from .50. This is barely off of the mean of the standard normal distribution of zero. There is virtually no difference from the sample proportion and the hypothesized proportion in terms of standard deviations.

STEP 4 : Compare the test statistic and the critical value.

The calculated value is well within the critical values of ± 1.96 standard deviations and thus we cannot reject the null hypothesis. To reject the null hypothesis we need significant evident of difference between the hypothesized value and the sample value. In this case the sample value is very nearly the same as the hypothesized value measured in terms of standard deviations.

STEP 5 : Reach a conclusion

The formal conclusion would be “At a 5% level of significance we cannot reject the null hypothesis that 50% of first-time borrowers take out smaller loans than other borrowers.” Notice the length to which the conclusion goes to include all of the conditions that are attached to the conclusion. Statisticians, for all the criticism they receive, are careful to be very specific even when this seems trivial. Statisticians cannot say more than they know, and the data constrain the conclusion to be within the metes and bounds of the data.

Try It 9.11

A teacher believes that 85% of students in the class will want to go on a field trip to the local zoo. The teacher performs a hypothesis test to determine if the percentage is the same or different from 85%. The teacher samples 50 students and 39 reply that they would want to go to the zoo. For the hypothesis test, use a 1% level of significance.

Example 9.12

Suppose a consumer group suspects that the proportion of households that have three or more cell phones is 30%. A cell phone company has reason to believe that the proportion is not 30%. Before they start a big advertising campaign, they conduct a hypothesis test. Their marketing people survey 150 households with the result that 43 of the households have three or more cell phones.

Here is an abbreviate version of the system to solve hypothesis tests applied to a test on a proportions.

Try It 9.12

Marketers believe that 92% of adults in the United States own a cell phone. A cell phone manufacturer believes that number is actually lower. 200 American adults are surveyed, of which, 174 report having cell phones. Use a 5% level of significance. State the null and alternative hypothesis, find the p -value, state your conclusion, and identify the Type I and Type II errors.

Example 9.13

The National Institute of Standards and Technology provides exact data on conductivity properties of materials. Following are conductivity measurements for 11 randomly selected pieces of a particular type of glass.

1.11; 1.07; 1.11; 1.07; 1.12; 1.08; .98; .98; 1.02; .95; .95 Is there convincing evidence that the average conductivity of this type of glass is greater than one? Use a significance level of 0.05.

Let’s follow a four-step process to answer this statistical question.

• H 0 : μ ≤ 1
• H a : μ > 1
• Plan : We are testing a sample mean without a known population standard deviation with less than 30 observations. Therefore, we need to use a Student's-t distribution. Assume the underlying population is normal.
• Do the calculations and draw the graph .
• State the Conclusions : We cannot accept the null hypothesis. It is reasonable to state that the data supports the claim that the average conductivity level is greater than one.

Try It 9.13

The boiling point of a specific liquid is measured for 15 samples, and the boiling points are obtained as follows:

205; 206; 206; 202; 199; 194; 197; 198; 198; 201; 201; 202; 207; 211; 205

Is there convincing evidence that the average boiling point is greater than 200? Use a significance level of 0.1. Assume the population is normal.

Example 9.14

In a study of 420,019 cell phone users, 172 of the subjects developed brain cancer. Test the claim that cell phone users developed brain cancer at a greater rate than that for non-cell phone users (the rate of brain cancer for non-cell phone users is 0.0340%). Since this is a critical issue, use a 0.005 significance level. Explain why the significance level should be so low in terms of a Type I error.

• H 0 : p ≤ 0.00034
• H a : p > 0.00034

If we commit a Type I error, we are essentially accepting a false claim. Since the claim describes cancer-causing environments, we want to minimize the chances of incorrectly identifying causes of cancer.

• We will be testing a sample proportion with x = 172 and n = 420,019. The sample is sufficiently large because we have np' = 420,019(0.00034) = 142.8, nq' = 420,019(0.99966) = 419,876.2, two independent outcomes, and a fixed probability of success p' = 0.00034. Thus we will be able to generalize our results to the population.

Try It 9.14

In a study of 390,000 moisturizer users, 138 of the subjects developed skin diseases. Test the claim that moisturizer users developed skin diseases at a greater rate than that for non-moisturizer users (the rate of skin diseases for non-moisturizer users is 0.041%). Since this is a critical issue, use a 0.005 significance level. Explain why the significance level should be so low in terms of a Type I error.

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Business Statistics with Solutions in R

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• Copyright year: 2019
• Audience: Policy makers, professionals in business statistics, applied statistics, statistical computing, finance, econometrics.
• Front matter: 12
• Main content: 264
• Illustrations: 30
• Keywords: Probability ; Sampling Distributions ; Confidence interval ; Regression ; Hypothesis testing ; Time series ; Expected value ; Correlation
• Published: October 21, 2019
• ISBN: 9781547401352
• ISBN: 9781547417469

Excel 2019 for Business Statistics

A Guide to Solving Practical Problems

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Newly revised to specifically provide demonstration in Excel 2019, this volume shows the capabilities of Microsoft Excel in business statistics. Similar to its predecessor, Excel 2016 for Business Statistics , it is a step-by-step, exercise-driven guide for students and practitioners who are looking to master Excel to solve practical business problems.

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Table of contents (8 chapters)

Front matter, sample size, mean, standard deviation, and standard error of the mean.

Thomas J. Quirk

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Confidence interval about the mean using the tinv function and hypothesis testing, one-group t-test for the mean, two-group t-test of the difference of the means for independent groups, correlation and simple linear regression, multiple correlation and multiple regression, one-way analysis of variance (anova), back matter, authors and affiliations, about the author.

At the beginning of his academic career, Thomas J. Quirk spent six years in educational research at The American Institutes for Research and Educational Testing Service. He is currently a Professor Emeritus of Marketing in The Walker School of Business & Technology at Webster University based in St. Louis, Missouri (USA) where he taught Marketing Statistics, Marketing Research, and Pricing Strategies. He has written 60+ textbook supplements in Marketing and Management, published 20+ articles in professional journals, and presented 20+ papers at professional meetings. He holds a BS in Mathematics from John Carroll University, both an MA in Education and a PhD in Educational Psychology from Stanford University, and an MBA from The University of Missouri-St. Louis.

Number Line

• arithmetic\:mean\:1,\:2,\:3,\:4,\:5,\:6
• geometric\:mean\:\left\{0.42,\:0.52,\:0.58,\:0.62\right\}
• quadratic\:mean\:-4,\:5,\:6,\:9
• median\:\:\left\{1,\:7,\:-3,\:4,\:9\right\}
• mode\:\left\{90,\:94,\:53,\:68,\:79,\:94,\:87,\:90,\:70,\:69,\:65,\:89,\:85\right\}
• minimum\:-4,\:5,\:6,\:9
• maximum\:\frac{31}{100},\:\frac{23}{105},\:\frac{31}{205},\:\frac{54}{205}
• mid\:range\:1,\:2,\:3,\:4,\:5,\:6
• range\:\:\left\{1,\:7,\:-3,\:4,\:9\right\}
• standard\:deviation\:\:\left\{1,\:7,\:-3,\:4,\:9\right\}
• variance\:1,\:2,\:3,\:4,\:5,\:6
• lower\:quartile\:-4,\:5,\:6,\:9
• upper\:quartile\:\left\{0.42,\:0.52,\:0.58,\:0.62\right\}
• interquartile\:range\:1,\:2,\:3,\:4,\:5,\:6
• midhinge\:\left\{90,\:94,\:53,\:68,\:79,\:84,\:87,\:72,\:70,\:69,\:65,\:89,\:85\right\}
• What is the best calculator for statistics?
• Symbolab offers an online calculator specifically for statistics that can perform a wide range of calculations, including standard deviation, variance, range and normal distribution. It also provides detailed step-by-step solutions.
• What is statistics?
• Statistics is the branch of mathematics that deals with the collection, analysis, interpretation, presentation, and organization of data. There are two main branches of statistics: descriptive statistics, and inferential statistics.
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• Descriptive statistics is a branch of statistics that deals with summarizing, organizing and describing data. Descriptive statistics uses measures such as central tendency (mean, median, and mode) and measures of variability (range, standard deviation, variance) to give an overview of the data.
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• Inferential statistics is a branch of statistics that deals with making predictions and inferences about a population based on a sample of data. Inferential statistics uses probability theory and statistical models to make predictions and inferences about a population.
• What is the difference between statistics and probability?
• Statistics is the branch of mathematics dealing with the collection, analysis, interpretation, presentation, and organization of data, while probability is the branch of mathematics dealing with the likelihood of occurrence of different events.

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What Is Regression Analysis in Business Analytics?

• 14 Dec 2021

Countless factors impact every facet of business. How can you consider those factors and know their true impact?

Imagine you seek to understand the factors that influence people’s decision to buy your company’s product. They range from customers’ physical locations to satisfaction levels among sales representatives to your competitors' Black Friday sales.

Understanding the relationships between each factor and product sales can enable you to pinpoint areas for improvement, helping you drive more sales.

To learn how each factor influences sales, you need to use a statistical analysis method called regression analysis .

If you aren’t a business or data analyst, you may not run regressions yourself, but knowing how analysis works can provide important insight into which factors impact product sales and, thus, which are worth improving.

Access your free e-book today.

Foundational Concepts for Regression Analysis

Before diving into regression analysis, you need to build foundational knowledge of statistical concepts and relationships.

Independent and Dependent Variables

Start with the basics. What relationship are you aiming to explore? Try formatting your answer like this: “I want to understand the impact of [the independent variable] on [the dependent variable].”

The independent variable is the factor that could impact the dependent variable . For example, “I want to understand the impact of employee satisfaction on product sales.”

In this case, employee satisfaction is the independent variable, and product sales is the dependent variable. Identifying the dependent and independent variables is the first step toward regression analysis.

Correlation vs. Causation

One of the cardinal rules of statistically exploring relationships is to never assume correlation implies causation. In other words, just because two variables move in the same direction doesn’t mean one caused the other to occur.

If two or more variables are correlated , their directional movements are related. If two variables are positively correlated , it means that as one goes up or down, so does the other. Alternatively, if two variables are negatively correlated , one goes up while the other goes down.

A correlation’s strength can be quantified by calculating the correlation coefficient , sometimes represented by r . The correlation coefficient falls between negative one and positive one.

r = -1 indicates a perfect negative correlation.

r = 1 indicates a perfect positive correlation.

r = 0 indicates no correlation.

Causation means that one variable caused the other to occur. Proving a causal relationship between variables requires a true experiment with a control group (which doesn’t receive the independent variable) and an experimental group (which receives the independent variable).

While regression analysis provides insights into relationships between variables, it doesn’t prove causation. It can be tempting to assume that one variable caused the other—especially if you want it to be true—which is why you need to keep this in mind any time you run regressions or analyze relationships between variables.

With the basics under your belt, here’s a deeper explanation of regression analysis so you can leverage it to drive strategic planning and decision-making.

Related: How to Learn Business Analytics without a Business Background

What Is Regression Analysis?

Regression analysis is the statistical method used to determine the structure of a relationship between two variables (single linear regression) or three or more variables (multiple regression).

According to the Harvard Business School Online course Business Analytics , regression is used for two primary purposes:

• To study the magnitude and structure of the relationship between variables
• To forecast a variable based on its relationship with another variable

Both of these insights can inform strategic business decisions.

“Regression allows us to gain insights into the structure of that relationship and provides measures of how well the data fit that relationship,” says HBS Professor Jan Hammond, who teaches Business Analytics, one of three courses that comprise the Credential of Readiness (CORe) program . “Such insights can prove extremely valuable for analyzing historical trends and developing forecasts.”

One way to think of regression is by visualizing a scatter plot of your data with the independent variable on the X-axis and the dependent variable on the Y-axis. The regression line is the line that best fits the scatter plot data. The regression equation represents the line’s slope and the relationship between the two variables, along with an estimation of error.

Physically creating this scatter plot can be a natural starting point for parsing out the relationships between variables.

Types of Regression Analysis

There are two types of regression analysis: single variable linear regression and multiple regression.

Single variable linear regression is used to determine the relationship between two variables: the independent and dependent. The equation for a single variable linear regression looks like this:

In the equation:

• ŷ is the expected value of Y (the dependent variable) for a given value of X (the independent variable).
• x is the independent variable.
• α is the Y-intercept, the point at which the regression line intersects with the vertical axis.
• β is the slope of the regression line, or the average change in the dependent variable as the independent variable increases by one.
• ε is the error term, equal to Y – ŷ, or the difference between the actual value of the dependent variable and its expected value.

Multiple regression , on the other hand, is used to determine the relationship between three or more variables: the dependent variable and at least two independent variables. The multiple regression equation looks complex but is similar to the single variable linear regression equation:

Each component of this equation represents the same thing as in the previous equation, with the addition of the subscript k, which is the total number of independent variables being examined. For each independent variable you include in the regression, multiply the slope of the regression line by the value of the independent variable, and add it to the rest of the equation.

How to Run Regressions

You can use a host of statistical programs—such as Microsoft Excel, SPSS, and STATA—to run both single variable linear and multiple regressions. If you’re interested in hands-on practice with this skill, Business Analytics teaches learners how to create scatter plots and run regressions in Microsoft Excel, as well as make sense of the output and use it to drive business decisions.

Calculating Confidence and Accounting for Error

It’s important to note: This overview of regression analysis is introductory and doesn’t delve into calculations of confidence level, significance, variance, and error. When working in a statistical program, these calculations may be provided or require that you implement a function. When conducting regression analysis, these metrics are important for gauging how significant your results are and how much importance to place on them.

Why Use Regression Analysis?

Once you’ve generated a regression equation for a set of variables, you effectively have a roadmap for the relationship between your independent and dependent variables. If you input a specific X value into the equation, you can see the expected Y value.

This can be critical for predicting the outcome of potential changes, allowing you to ask, “What would happen if this factor changed by a specific amount?”

Returning to the earlier example, running a regression analysis could allow you to find the equation representing the relationship between employee satisfaction and product sales. You could input a higher level of employee satisfaction and see how sales might change accordingly. This information could lead to improved working conditions for employees, backed by data that shows the tie between high employee satisfaction and sales.

Whether predicting future outcomes, determining areas for improvement, or identifying relationships between seemingly unconnected variables, understanding regression analysis can enable you to craft data-driven strategies and determine the best course of action with all factors in mind.

Do you want to become a data-driven professional? Explore our eight-week Business Analytics course and our three-course Credential of Readiness (CORe) program to deepen your analytical skills and apply them to real-world business problems.

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Statistics for Business Problem Solving 2nd Edition

• ISBN-10 9780538831307
• ISBN-13 978-0538831307
• Edition 2nd
• Publisher South-Western College Pub
• Publication date July 22, 1994
• Language English
• Dimensions 1.5 x 8.5 x 10.5 inches
• Print length 704 pages
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Product details

• ASIN ‏ : ‎ 0538831308
• Publisher ‏ : ‎ South-Western College Pub; 2nd edition (July 22, 1994)
• Language ‏ : ‎ English
• Hardcover ‏ : ‎ 704 pages
• ISBN-10 ‏ : ‎ 9780538831307
• ISBN-13 ‏ : ‎ 978-0538831307
• Item Weight ‏ : ‎ 4.25 pounds
• Dimensions ‏ : ‎ 1.5 x 8.5 x 10.5 inches
• #3,487 in Business Statistics
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About the author

Harvey j. brightman.

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Effective Problem-Solving Techniques in Business

Problem solving is an increasingly important soft skill for those in business. The Future of Jobs Survey by the World Economic Forum drives this point home. According to this report, complex problem solving is identified as one of the top 15 skills that will be sought by employers in 2025, along with other soft skills such as analytical thinking, creativity and leadership.

Dr. Amy David , clinical associate professor of management for supply chain and operations management, spoke about business problem-solving methods and how the Purdue University Online MBA program prepares students to be business decision-makers.

Why Are Problem-Solving Skills Essential in Leadership Roles?

Every business will face challenges at some point. Those that are successful will have people in place who can identify and solve problems before the damage is done.

“The business world is constantly changing, and companies need to be able to adapt well in order to produce good results and meet the needs of their customers,” David says. “They also need to keep in mind the triple bottom line of ‘people, profit and planet.’ And these priorities are constantly evolving.”

To that end, David says people in management or leadership need to be able to handle new situations, something that may be outside the scope of their everyday work.

“The name of the game these days is change—and the speed of change—and that means solving new problems on a daily basis,” she says.

The pace of information and technology has also empowered the customer in a new way that provides challenges—or opportunities—for businesses to respond.

“Our customers have a lot more information and a lot more power,” she says. “If you think about somebody having an unhappy experience and tweeting about it, that’s very different from maybe 15 years ago. Back then, if you had a bad experience with a product, you might grumble about it to one or two people.”

David says that this reality changes how quickly organizations need to react and respond to their customers. And taking prompt and decisive action requires solid problem-solving skills.

What Are Some of the Most Effective Problem-Solving Methods?

David says there are a few things to consider when encountering a challenge in business.

“When faced with a problem, are we talking about something that is broad and affects a lot of people? Or is it something that affects a select few? Depending on the issue and situation, you’ll need to use different types of problem-solving strategies,” she says.

Using Techniques

There are a number of techniques that businesses use to problem solve. These can include:

• Five Whys : This approach is helpful when the problem at hand is clear but the underlying causes are less so. By asking “Why?” five times, the final answer should get at the potential root of the problem and perhaps yield a solution.
• Gap Analysis : Companies use gap analyses to compare current performance with expected or desired performance, which will help a company determine how to use its resources differently or adjust expectations.
• Gemba Walk : The name, which is derived from a Japanese word meaning “the real place,” refers to a commonly used technique that allows managers to see what works (and what doesn’t) from the ground up. This is an opportunity for managers to focus on the fundamental elements of the process, identify where the value stream is and determine areas that could use improvement.
• Porter’s Five Forces : Developed by Harvard Business School professor Michael E. Porter, applying the Five Forces is a way for companies to identify competitors for their business or services, and determine how the organization can adjust to stay ahead of the game.
• Six Thinking Hats : In his book of the same name, Dr. Edward de Bono details this method that encourages parallel thinking and attempting to solve a problem by trying on different “thinking hats.” Each color hat signifies a different approach that can be utilized in the problem-solving process, ranging from logic to feelings to creativity and beyond. This method allows organizations to view problems from different angles and perspectives.
• SWOT Analysis : This common strategic planning and management tool helps businesses identify strengths, weaknesses, opportunities and threats (SWOT).

“We have a lot of these different tools,” David says. “Which one to use when is going to be dependent on the problem itself, the level of the stakeholders, the number of different stakeholder groups and so on.”

Each of the techniques outlined above uses the same core steps of problem solving:

• Identify and define the problem
• Consider possible solutions
• Evaluate options
• Choose the best solution
• Implement the solution
• Evaluate the outcome

Data drives a lot of daily decisions in business and beyond. Analytics have also been deployed to problem solve.

“We have specific classes around storytelling with data and how you convince your audience to understand what the data is,” David says. “Your audience has to trust the data, and only then can you use it for real decision-making.”

Data can be a powerful tool for identifying larger trends and making informed decisions when it’s clearly understood and communicated. It’s also vital for performance monitoring and optimization.

How Is Problem Solving Prioritized in Purdue’s Online MBA?

The courses in the Purdue Online MBA program teach problem-solving methods to students, keeping them up to date with the latest techniques and allowing them to apply their knowledge to business-related scenarios.

“I can give you a model or a tool, but most of the time, a real-world situation is going to be a lot messier and more valuable than what we’ve seen in a textbook,” David says. “Asking students to take what they know and apply it to a case where there’s not one single correct answer is a big part of the learning experience.”

Make Your Own Decision to Further Your Career

An online MBA from Purdue University can help advance your career by teaching you problem-solving skills, decision-making strategies and more. Reach out today to learn more about earning an online MBA with Purdue University .

If you would like to receive more information about pursuing a business master’s at the Mitchell E. Daniels, Jr. School of Business, please fill out the form and a program specialist will be in touch!

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