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A company owned two small Bath and Body Goods stores in different cities. It was desired to see if there was a difference in their mean daily sales. The following results were obtained from a random sample of daily sales over a six-week period. At \(\alpha=0.01,\) can a difference in sales be concluded? Use the \(P\) -value method. $$ \begin{array}{llcc} & & \text { Population } & \\ & & \text { standard } & \text { Sample } \\ \text { Store } & \text { Mean } & \text { deviation } & \text { size } \\ \hline \text { A } & \$ 995 & \$ 120 & 30 \\ \text { B } & 1120 & 250 & 30 \end{array} $$

Short Answer

Expert verified
There is no evidence to conclude a difference in sales since the P-value is greater than 0.01.

Step by step solution

01

Formulate Hypotheses

Define the null hypothesis as \(H_0: \mu_A = \mu_B\) (no difference in means) and the alternative hypothesis as \(H_1: \mu_A eq \mu_B\) (a difference in means).
02

Identify the Significance Level

The significance level \(\alpha\) is given as 0.01.
03

Calculate the Test Statistic

Use the formula for the test statistic for the difference in means: \[ z = \frac{(\bar{x}_A - \bar{x}_B)}{\sqrt{\frac{\sigma_A^2}{n_A} + \frac{\sigma_B^2}{n_B}}} \] where \(\bar{x}_A = 995\), \(\bar{x}_B = 1120\), \(\sigma_A = 120\), \(\sigma_B = 250\), \(n_A = 30\), and \(n_B = 30\). Plug in the values to get: \[ z = \frac{(995 - 1120)}{\sqrt{\frac{120^2}{30} + \frac{250^2}{30}}} \]
04

Compute the P-value

Calculate the P-value based on the computed Z-value. For a two-tailed test, find the probability that a Z-score is as extreme as the observed value in both tails.
05

Make Conclusions

Compare the P-value with \(\alpha = 0.01\). If the P-value is less than \(\alpha\), reject the null hypothesis \(H_0\). Otherwise, fail to reject \(H_0\).

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Key Concepts

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

p-value method
The p-value method helps you determine the strength of your evidence against the null hypothesis. When you calculate a test statistic based on your sample data, the p-value tells you how likely your observed results would occur if the null hypothesis were actually true. It's like putting a number on the surprise factor of your results. The smaller the p-value, the more surprising your results are, and the greater the reason to doubt the null hypothesis.

For hypothesis testing, you compare the p-value against your significance level (b1). This tells you whether your results are statistically significant. In this context, the p-value is derived from the test statistic, which reflects the likelihood of observing your sample data under the null hypothesis. It helps decide if there is enough evidence to reject the null hypothesis.
significance level
The significance level, denoted by b1, is a threshold set by the researcher which dictates how much risk of making a Type I error (rejecting the true null hypothesis) they're willing to accept. It represents a probability, often set at 0.01, 0.05, or 0.10. In this exercise, the significance level is set at 0.01, illustrating a high standard of evidence.

When b1 is at 0.01, it means you're accepting only a 1% risk of committing a Type I error. In simpler terms, you want to be 99% confident about your decision to reject the null hypothesis. This high confidence requirement is often used in fields where making errors could have serious consequences.
  • Set before testing begins, preserving the integrity of your decision-making process.
  • Helps balance the risk of different types of errors in hypothesis testing.
test statistic calculation
Calculating the test statistic is a crucial step in hypothesis testing and involves using a formula that relates to the nature of your data and the hypotheses being tested. In this scenario, where you're comparing the means of two independent samples, the Z test statistic is appropriate, primarily due to the sample sizes and known population standard deviations.

The formula used is: \[ z = \frac{(\bar{x}_A - \bar{x}_B)}{\sqrt{\frac{\sigma_A^2}{n_A} + \frac{\sigma_B^2}{n_B}}} \] Where:
  • \(\bar{x}_A\) and \(\bar{x}_B\) are sample means,
  • \(\sigma_A\) and \(\sigma_B\) are population standard deviations,
  • \(n_A\) and \(n_B\) are sample sizes.
By substituting the given values into the formula, you compute a Z score that then allows you to determine the p-value.
null and alternative hypotheses
The null and alternative hypotheses are foundational to hypothesis testing. They form the basis of what you're trying to test and conclude.

- **Null Hypothesis (329;\(_0\))**: This hypothesis assumes that there is no effect or difference. It's the statement being tested, and your goal is to assess the evidence against it. In this problem, the null hypothesis is that the mean daily sales in both stores are equal, denoted as \(H_0: \mu_A = \mu_B\).

- **Alternative Hypothesis (\(H_1\))**: This is what you would think is true if the null hypothesis is rejected. It represents a new effect or difference that might be true. For the current exercise, the alternative hypothesis claims there is a difference in the means of daily sales between the two stores, expressed as \(H_1: \mu_A eq \mu_B\).
  • The null hypothesis is like the status quo, assuming no change or difference.
  • The alternative hypothesis is what you're looking to prove, indicating a significant effect or difference.
Choosing the correct hypotheses is critical, as it influences the direction and nature of your statistical testing.

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Most popular questions from this chapter

A researcher claims that students in a private school have exam scores that are at most 8 points higher than those of students in public schools. Random samples of 60 students from each type of school are selected and given an exam. The results are shown. At \(\alpha=0.05,\) test the claim. $$ \begin{array}{cc} \text { Private school } & \text { Public school } \\ \hline \bar{X}_{1}=110 & \bar{X}_{2}=104 \\ \sigma_{1}=15 & \sigma_{2}=15 \\ n_{1}=60 & n_{2}=60 \end{array} $$

Perform these steps. a. State the hypotheses and identify the claim. b. Find the critical value(s). c. Compute the test value. d. Make the decision. e. Summarize the results. Use the traditional method of hypothesis testing unless otherwise specified. The percentage of males 18 years and older who have never married is \(30.4 .\) For females the percentage is \(23.6 .\) Looking at the records in a particular populous county, a random sample of 250 men showed that 78 had never married and 58 of 200 women had never married. At the 0.05 level of significance, is the proportion of men greater than the proportion of women? Use the \(P\) -value method.

Perform these steps. a. State the hypotheses and identify the claim. b. Find the critical value(s). c. Compute the test value. d. Make the decision. e. Summarize the results. Use the traditional method of hypothesis testing unless otherwise specified. In a random sample of 200 men, 130 said they used seat belts. In a random sample of 300 women, 63 said they used seat belts. Test the claim that men are more safety-conscious than women, at \(\alpha=0.01\). Use the \(P\) -value method.

Perform these steps. a. State the hypotheses and identify the claim. b. Find the critical value(s). c. Compute the test value. d. Make the decision. e. Summarize the results. Use the traditional method of hypothesis testing unless otherwise specified. It has been found that many firsttime interviewees commit errors that could very well affect the outcome of the interview. An astounding \(77 \%\) are guilty of using their cell phones or texting during the interview! A researcher wanted to see if the proportion of male offenders differed from the proportion of female ones. Out of 120 males, 72 used their cell phone and 80 of 150 females did so. At the 0.01 level of significance is there a difference?

Perform each of these steps. Assume that all variables are normally or approximately normally distributed a. State the hypotheses and identify the claim. b. Find the critical value(s). c. Compute the test value. d. Make the decision. e. Summarize the results. Use the traditional method of hypothesis testing unless otherwise specified. Obstacle Course Times An obstacle course was set up on a campus, and 8 randomly selected volunteers were given a chance to complete it while they were being timed. They then sampled a new energy drink and were given the opportunity to run the course again. The "before" and "after" times in seconds are shown. Is there sufficient evidence at \(\alpha=0.05\) to conclude that the students did better the second time? Discuss possible reasons for your results. $$ \begin{array}{l|rrrrrrrr} \text { Student } & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 \\ \hline \text { Before } & 67 & 72 & 80 & 70 & 78 & 82 & 69 & 75 \\ \hline \text { After } & 68 & 70 & 76 & 65 & 75 & 78 & 65 & 68 \end{array} $$

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