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Chapter 10: Problem 47

Web business You want to compare the daily sales for two different designs of Web pages for your Internet business. You assign the next 60 days to either Design A or Design B, 30 days to each. (a) Describe how you would assign the days for Design A and Design B using the partial line of random digits provided below. Then use your plan to select the first three days for using Design A. Show your method clearly on your paper. $$24005 \qquad 52114 \qquad 26224 \qquad 39078$$ (b) Would you use a one-sided or a two-sided significance test for this problem? Explain your choice. Then set up appropriate hypotheses. (c) If you plan to use Table B to calculate the P-value, what are the degrees of freedom? (d) The t statistic for comparing the mean sales is 2.06. Using Table B, what P-value would you report? What would you conclude?

Short Answer

Expert verified
The first three days for Design A are Day 1, Day 2, and Day 3. Use a two-sided test; degrees of freedom are 58. P-value shows a significant difference.

Step by step solution

01

Assigning Days Using Random Digits

To assign 30 days to each design, read the random digits in pairs. Assign 'Design A' to even pairs (00-98 where both digits are even) and 'Design B' to odd pairs. Starting from the left, '24' (Design A), '00' (Design A), and '22' (Design A) are the first three pairs, so the first three days are assigned to Design A.
02

Choosing the Significance Test Type

A two-sided significance test is appropriate because you are interested in comparing the sales performance of both designs without specifically focusing on which design might result in higher or lower sales. This means you need to determine if there is any difference, not just an increase or decrease.
03

Setting Up Hypotheses

The null hypothesis (H0) is that there is no difference in the mean sales for Design A and Design B: \( H_0: \mu_A = \mu_B \). The alternative hypothesis (H1) is that there is a difference: \( H_1: \mu_A eq \mu_B \).
04

Determining Degrees of Freedom

Since each design has 30 days of data, the degrees of freedom for the P-value calculation in a two-sample t-test is \( n_A + n_B - 2 = 30 + 30 - 2 = 58 \).
05

Calculating the P-value and Drawing Conclusions

With a t statistic of 2.06 and 58 degrees of freedom, use Table B to find the P-value. This P-value is slightly less than 0.05, indicating a statistically significant difference at the 5% level. Thus, we reject the null hypothesis, concluding there is a significant difference between the sales from the two designs.

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Key Concepts

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

Random Assignment
Random assignment is a fundamental method in experimental research used to ensure each participant or, in this scenario, each day, has an equal chance of being assigned to either Design A or Design B. This technique minimizes biases and external influences, giving more reliable results. Let's break down how this was applied in the exercise.

- **Using Random Digits:** The exercise uses a line of random digits to assign 30 days to Design A and 30 days to Design B.
- **Pair Reading:** You read the numbers in pairs from the random digit line. Even pairs, such as '24', '00', and '22', are assigned to Design A. Odd pairs would belong to Design B.

This method assures that the assignment of days to each design is free of personal bias and random in nature.
Two-sided Significance Test
When comparing two different designs, such as Design A and Design B, a two-sided significance test is often chosen. This is because the goal is to find out if there is any significant difference in sales, regardless of whether one is higher or lower. Here鈥檚 why this approach is suitable for the exercise.

- **Objective of Testing:** A two-sided test checks for any deviations from the null hypothesis in either direction. It looks for any differences in mean sales, rather than just an increase or decrease.
- **Why Two-sided:** In the scenario, you're not just interested in knowing if one design performs better, but if they have different effects at all.

Hence, by using a two-sided test, you ensure thoroughness, capturing any potential difference in sales performance between the two designs.
Hypothesis Testing
Hypothesis testing in statistics allows us to make decisions or inferences about population parameters based on sample data. In this exercise, you'll perform hypothesis testing on the sales data for Design A and Design B. Let's explore the setup here:

- **Null Hypothesis (H0):** This assumes there's no difference in mean sales between Design A and Design B, formally written as \( H_0: \mu_A = \mu_B \).
- **Alternative Hypothesis (H1):** It suggests that there is a difference in the means, written as \( H_1: \mu_A eq \mu_B \).

The results of hypothesis testing will guide whether we reject the null hypothesis, implying a significant difference in sales between the two web page designs.
Degrees of Freedom
Degrees of freedom play an essential role in the context of statistical tests by indicating the number of values in a calculation that are free to vary. When comparing the means of two groups, as in this exercise, the degrees of freedom are calculated as follows:

- **Calculation:** For two groups, each with 30 days of data, the formula is \( n_A + n_B - 2 \). With \( n_A = 30 \) and \( n_B = 30 \), you have \( 58 \) degrees of freedom.

- **Why Important:** Degrees of freedom help determine the appropriate distribution to use when calculating the P-value for the t-test. It affects the shape of the t-distribution, which is needed when finding out whether observed differences in means are statistically significant.

Understanding degrees of freedom ensures you correctly apply statistical models to get valid results.

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Most popular questions from this chapter

Exercises 71 to 74 refer to the following setting. Coaching companies claim that their courses can raise the SAT scores of high school students. Of course, students who retake the SAT without paying for coaching generally raise their scores. A random sample of students who took the SAT twice found 427 who were coached and 2733 who were uncoached.\(^{44}\) Starting with their Verbal scores on the first and second tries, we have these summary statistics: Coaching and SAT scores (10.1) What proportion of students who take the SAT twice are coached? To answer this question, Jannie decides to construct a 99% confidence interval. Her work is shown below. Explain what鈥檚 wrong with Jannie鈥檚 method. $$hat{p}_{1}=\frac{427}{3160}=0.135=\underset{\text { who were coached }}{\text { proportion of students }}$$ $$\hat{p}_{2}=\frac{2733}{3160}=0.865=\underset{\text { who weren't coached }}{\text { who weren't coached }}$$ $$\mathrm{A} 99 \% \mathrm{Cl} \text { for } p_{1}-p_{2} \mathrm{is}$$ $$\begin{aligned}(0.135-0.865) \pm 2.575 \sqrt{\frac{0.135(0.865)}{3160}} &+\frac{0.865(0.135)}{2733} \\ &=-0.73 \pm 0.022=(-0.752,-0.708) \end{aligned}$$ We are 99% confident that the proportion of students taking the SAT twice who are coached is between 71 and 75 percentage points lower than students who aren鈥檛 coached.

Explain why the conditions for using two-sample z procedures to perform inference about \(p_{1}-p_{2}\) are not met in the settings of Exercises 7 through 10 . Don鈥檛 drink the water! The movie A Civil Action (Touchstone Pictures, 1998) tells the story of a major legal battle that took place in the small town of Woburn, Massachusetts. A town well that supplied water to eastern Woburn residents was contaminated by industrial chemicals. During the period that residents drank water from this well, 16 of the 414 babies born had birth defects. On the west side of Woburn, 3 of the 228 babies born during the same time period had birth defects.

What鈥檚 wrong? 鈥淲ould you marry a person from a lower social class than your own?鈥� Researchers asked this question of a random sample of 385 black, never- married students at two historically black colleges in the South. Of the 149 men in the sample, 91 said 鈥淵es.鈥� Among the 236 women, 117 said 鈥淵es.鈥漒(^{14}\) Is there reason to think that different proportions of men and women in this student population would be willing to marry beneath their class? Holly carried out the significance test shown below to answer this question. Unfortunately, she made some mistakes along the way. Identify as many mistakes as you can, and tell how to correct each one. State: I want to perform a test of $$H_{0} : p_{1}=\rho_{2}$$ $$H_{a} : p_{1} \neq p_{2}$$ at the 95% confidence level. Plan: If conditions are met, I鈥檒l do a one-sample \(z\) test for comparing two proportions. \(\cdot\) Random The data came from a random sample of 385 black, never-married students. \(\cdot\) Normal One student's answer to the question should have no relationship to another student's answer. \(\cdot\) Independent The counts of successes and falures in the two groups - \(91,58,117,\) and \(119-\) are all at least 10 . Do: From the data, \(\hat{p}_{1}=\frac{91}{149}=0.61\) and \(\hat{p}_{2}=\frac{117}{236}=0.46\) \(\bullet\) Test statistic $$z=\frac{(0.61-0.46)-0}{\sqrt{\frac{0.61(0.39)}{149}+\frac{0.46(0.54)}{236}}}=2.91$$ \(\cdot P\) value From Table \(A, P(z \geq 2.91)=1-0.9982=\) 0.0018 . Conclude: The P-value, \(0.0018,\) is less than \(0.05,\) so I'll reject the null hypothesis. This proves that a higher proportion of men than women are willing to marry someone from a social class lower than their own.

Is red wine better than white wine? Observational studies suggest that moderate use of alcohol by adults reduces heart attacks and that red wine may have special benefits. One reason may be that red wine contains polyphenols, substances that do good things to cholesterol in the blood and so may reduce the risk of heart attacks. In an experiment, healthy men were assigned at random to drink half a bottle of either red or white wine each day for two weeks. The level of polyphenols in their blood was measured before and after the two-week period. Here are the percent changes in level for the subjects in both groups:\(^{31}\) Red wine: 3.58 .1\(\quad 7.44 .0 \quad 0.74 .98 .4 \quad 7.05 .5\) White wine: \(3.1 \quad 0.5-3.8\) 4.1 \(-0.62 .7 \quad 1.9-5.9 \quad 0.1\) (a) A Fathom dotplot of the data is shown below. Use the graph to answer these questions: \(\bullet\) Are the centers of the two groups similar or different? Explain. \(\bullet\) Are the spreads of the two groups similar or different? Explain. (b) Construct and interpret a 90% confidence interval for the difference in mean percent change in polyphenol levels for the red wine and white wine treatments. (c) Does the interval in part (b) suggest that red wine is more effective than white wine? Explain.

Multiple choice: Select the best answer for Exercises 29 to 32. A sample survey interviews SRSs of 500 female college students and 550 male college students. Each student is asked whether he or she worked for pay last summer. In all, 410 of the women and 484 of the men say 鈥淵es.鈥� Exercises 29 to 31 are based on this survey. Take \(\rho_{M}\) and \(p_{F}\) to be the proportions of all college males and females who worked last summer. We conjectured before seeing the data that men are more likely to work. The hypotheses to be tested are (a) \(H_{0} : p_{M}-p_{F}=0\) versus \(H_{a} : p_{M}-p_{F} \neq 0\) (b) \(H_{0} : p_{M}-p_{F}=0\) versus \(H_{a} : p_{M}-p_{F}>0\) (c) \(H_{0} : p_{M}-p_{P}=0\) versus \(H_{a} : p_{M}-p_{F}<0\) (d) \(H_{0} : p_{M}-p_{P}>0\) versus \(H_{a} : p_{M}-p_{P}=0\) (e) \(H_{0} : p_{M}-p_{F} \neq 0\) versus \(H_{a} : p_{M}-p_{F}=0\)
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