/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 23 Exercise 13.16 described a regre... [FREE SOLUTION] | 91Ó°ÊÓ

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Exercise 13.16 described a regression analysis in which \(y=\) sales revenue and \(x=\) advertising expenditure. Summary quantities given there yield \(n=15 \quad b=52.27 \quad s_{b}=8.05\) a. Test the hypothesis \(H_{0}: \beta=0\) versus \(H_{x}: \beta \neq 0\) using a significance level of .05. What does your conclusion say about the nature of the relationship between \(x\) and \(y\) ? b. Consider the hypothesis \(H_{0}: \beta=40\) versus \(H_{A} \cdot \beta>\) 40\. The null hypothesis states that the average change in sales revenue associated with a 1 -unit increase in advertising expenditure is (at most) \(\$ 40,000\). Carry out a test using significance level .01 .

Short Answer

Expert verified
a. There is a statistically significant relationship between advertising expenditure and sales revenue. b. There isn't enough evidence to conclude that a unit increase in advertising expenditure increases sales revenue by more than $40,000.

Step by step solution

01

Testing the Hypothesis \(H_{0}: \beta=0\) against \(H_{1}: \beta \neq 0\)

To test the null hypothesis \(H_{0}: \beta=0\), use the provided summary quantities. The test statistic is calculated by \(t=b/s_{b}\), substituting the given values, you get \(t = 52.27/8.05 ≈ 6.49\). Next, compare this calculated t-statistic with the critical t-value for a two-tailed test at .05 significance level with degrees of freedom (=n-2). For df=15-2=13, the critical t-value is approximately ±2.160368. The calculated t-statistic is greater than the critical value, and hence is in the rejection region. Therefore, we reject the null hypothesis.
02

Interpreting the Result for \(H_{0}: \beta=0\) against \(H_{1}: \beta \neq 0\)

Rejecting the null hypothesis implies there exists a statistically significant relationship between advertising expenditure and sales revenue. In other words, the results suggest that changes in advertising expenditure have a significant effect on sales revenue.
03

Testing the Hypothesis \(H_{0}: \beta=40\) against \(H_{1}: \beta > 40\)

To test the null hypothesis \(H_{0}: \beta=40\), you calculate the t-statistic using the formula \(t=(b- \beta ) / s_{b} = (52.27-40) / 8.05 ≈ 1.52\) Then, compare this calculated t-statistic with the critical t-value for a one-tailed test at .01 significance level with degrees of freedom (=n-2 = 13). The critical t-value is approximately 2.650. Here, the calculated t-statistic is less than the critical value and hence it is not in the rejection region. Therefore, we fail to reject the null hypothesis.
04

Interpreting the Result for \(H_{0}: \beta=40\) against \(H_{1}: \beta > 40\)

Failing to reject the null hypothesis implies there isn't enough evidence to conclude that the average change in sales revenue associated with a 1 -unit increase in advertising expenditure is more than $40,000. In other words, the data does not provide strong enough evidence to reject the claim that a 1-unit increase in advertising expenditure on average increases sales revenue by $40,000 or less.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Hypothesis Testing
Hypothesis testing is a method used to make decisions or inferences about population parameters based on sample data. In the problem shared, you're given two hypotheses to test. The first scenario tests the hypothesis that advertising expenditure (\( x \)) has no effect on sales revenue (\( y \)). This is stated as \( H_0: \beta = 0 \). The alternative hypothesis, \( H_a: \beta eq 0 \), suggests that there is a significant effect. The process involves:
  • Calculating a test statistic from your data.
  • Comparing the test statistic to a critical value from a statistical distribution (like t-distribution).
  • Deciding whether to reject or not reject the null hypothesis based on this comparison.
If the test statistic is beyond the critical value (in the rejection region), the null hypothesis is rejected, indicating a significant relationship. In our case, rejecting \( H_0 \) indicates that advertising spending indeed affects sales revenue.
T-distribution
The t-distribution is a type of probability distribution that is symmetric and bell-shaped, similar to the normal distribution, but with heavier tails. It's particularly useful when dealing with small sample sizes (typically \( n < 30 \)) and when the population standard deviation is unknown.

In the given exercise, since the sample size is 15, the t-distribution is used to determine critical values for hypothesis testing. The degrees of freedom (df) for the t-distribution are calculated as \( n-2 \) in a linear regression context, where \( n \) is 15, hence df = 13.
  • The critical t-value is derived from the t-table based on the chosen significance level.
  • The calculated t-statistic from sample data is compared to this critical value.
  • If the calculated t-statistic falls in the rejection region, we reject \( H_0 \).
Understanding the t-distribution helps ensure that you are using the correct approach to determine the significance of your results, especially when sample sizes are small.
Significance Level
The significance level, often denoted by \( \alpha \), is the probability of rejecting the null hypothesis when it is actually true. It's a threshold set by the researcher to control the level of risk they are willing to take in making a Type I error (false positive).
  • In the exercise, two significance levels are used: 0.05 and 0.01.
  • A 0.05 significance level means there is a 5% risk of concluding that a relationship exists when there isn't one.
  • At a 0.01 significance level, the criteria are stricter, allowing only a 1% risk of a Type I error.
Choosing the significance level is important as it directly influences the critical value from the t-distribution table. A lower significance level requires stronger evidence (a higher t-value) to reject the null hypothesis. In practical terms, it sets the confidence level for making predictions based on the data.
Advertising Expenditure
Advertising expenditure refers to the amount of money spent by a company to promote its products or services. In the context of regression analysis, it's often treated as an independent variable, with the hypothesis that changes in advertising budget can lead to changes in sales revenue.

This concept is central to the exercise, as the relationship between advertising expenditure and sales revenue is being tested. A significant finding could show that increasing advertising budgets leads to an increase in revenue, justifying higher investments.
  • It represents an investment aimed at boosting consumer awareness.
  • The measure of effectiveness is typically sales revenue, indicating the return on investment.
  • Understanding how advertising expenditure influences sales helps in strategic planning by allocating resources more efficiently.
Regression analysis for advertising decisions combines historical data and statistical methods to quantify the benefits of advertising investments quantitatively.
Sales Revenue
Sales revenue is the income a company generates from selling its goods or services before any costs or expenses are subtracted. In the regression framework of the given exercise, sales revenue is treated as a dependent variable affected by advertising expenditure.

This relationship highlights the financial impact of marketing decisions and serves as a metric for evaluating the effectiveness of advertising strategies.
  • Sales revenue is crucial for assessing the overall financial health and performance of a business.
  • It provides a measurable outcome to evaluate the return on advertising expenditures.
  • Analyzing fluctuations in sales revenue helps identify trends, enabling informed decision-making.
The effectiveness of an advertising campaign is often measured by its ability to influence sales revenue. A significant regression outcome suggests a positive return on the advertising investment, leading to strategic decisions in marketing spend allocations.

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Most popular questions from this chapter

Consider the accompanying data on \(x=\) advertising share and \(y=\) market share for a particular brand of soft drink during 10 randomly selected years. \(\begin{array}{lllll}x & .103 & .072 & .071 & .077 & .086 & .047 & .060 & .050 & .070 & .052\end{array}\) \(\begin{array}{lllll}y & .135 .125 & .120 & .086 & .079 & .076 & .065 & .059 & .051 & .039\end{array}\) a. Construct a scatterplot for these data. Do you think the simple linear regression model would be appropriate for describing the relationship between \(x\) and \(y\) ? b. Calculate the equation of the estimated regression line and use it to obtain the predicted market share when the advertising share is \(.09 .\) c. Compute \(r^{2}\). How would you interpret this value? d. Calculate a point estimate of \(\sigma .\) On how many degrees of freedom is your estimate based?

A random sample of \(n=347\) students was selected, and each one was asked to complete several questionnaires, from which a Coping Humor Scale value \(x\) and a Depression Scale value \(y\) were determined ("Depression and Sense of Humor." (Psychological Reports \([1994]: 1473-1474)\). The resulting value of the sample correlation coefficient was -.18 . a. The investigators reported that \(P\) -value \(<.05 .\) Do you agree? b. Is the sign of \(r\) consistent with your intuition? Explain. (Higher scale values correspond to more developed sense of humor and greater extent of depression.) c. Would the simple linear regression model give accurate predictions? Why or why not?

A sample of \(n=61\) penguin burrows was selected, and values of both \(y=\) trail length \((\mathrm{m})\) and \(x=\) soil hardness (force required to penetrate the substrate to a depth of \(12 \mathrm{~cm}\) with a certain gauge, in \(\mathrm{kg}\) ) were determined for each one ("Effects of Substrate on the Distribution of Magellanic Penguin Burrows," The Auk [1991]: \(923-933\) ). The equation of the least-squares line was \(\hat{y}=11.607-1.4187 x,\) and \(r^{2}=.386 .\) a. Does the relationship between soil hardness and trail length appear to be linear, with shorter trails associated with harder soil (as the article asserted)? Carry out an appropriate test of hypotheses. b. Using \(s_{\mathrm{e}}=2.35, \bar{x}=4.5,\) and \(\sum(x-\bar{x})^{2}=250,\) predict trail length when soil hardness is 6.0 in a way that conveys information about the reliability and precision of the prediction. c. Would you use the simple linear regression model to predict trail length when hardness is \(10.0 ?\) Explain your reasoning

Television is regarded by many as a prime culprit for the difficulty many students have in performing well in school. The article "The Impact of Athletics, PartTime Employment, and Other Activities on Academic Achievement" (Journal of College Student Development [1992]\(: 447-453)\) reported that for a random sample of \(n=528\) college students, the sample correlation coefficient between time spent watching television \((x)\) and grade point average \((y)\) was \(r=-.26\). a. Does this suggest that there is a negative correlation between these two variables in the population from which the 528 students were selected? Use a test with significance level .01 .

The shelf life of packaged food depends on many factors. Dry cereal is considered to be a moisturesensitive product (no one likes soggy cereal!) with the shelf life determined primarily by moisture content. In a study of the shelf life of one particular brand of cereal, \(x=\) time on shelf (days stored at \(73^{\circ} \mathrm{F}\) and \(50 \%\) relative humidity) and \(y=\) moisture content (\%) were recorded. The resulting data are from "Computer Simulation Speeds Shelf Life Assessments" (Package Engineering [1983]\(: 72-73)\). a. Summary quantities are $$ \sum x=269 \quad \sum y=51 \quad \sum x y=1081.5 $$ \(\sum y^{2}=190.78 \quad \sum x^{2}=7745\) Find the equation of the estimated regression line for predicting moisture content from time on the shelf. b. Does the simple linear regression model provide useful information for predicting moisture content from knowledge of shelf time? c. Find a \(95 \%\) interval for the moisture content of an individual box of cereal that has been on the shelf 30 days. d. According to the article, taste tests indicate that this brand of cereal is unacceptably soggy when the moisture content exceeds 4.1. Based on your interval in Part (c), do you think that a box of cereal that has been on the shelf 30 days will be acceptable? Explain.

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