/*! 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 26 Consider the following hypothesi... [FREE SOLUTION] | 91Ó°ÊÓ

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Consider the following hypothesis test: \\[ \begin{array}{l} H_{0}: \mu=100 \\ H_{\mathrm{a}}: \mu \neq 100 \end{array} \\] A sample of 65 is used. Identify the \(p\) -value and state your conclusion for each of the following sample results. Use \(\alpha=.05\) a. \(\bar{x}=103\) and \(s=11.5\) b. \(\bar{x}=96.5\) and \(s=11.0\) c. \(\bar{x}=102\) and \(s=10.5\)

Short Answer

Expert verified
a. Do not reject; b. Reject; c. Do not reject.

Step by step solution

01

Identify the Hypotheses and Significance Level

The null hypothesis \( H_0 \) is \( \mu = 100 \), and the alternative hypothesis \( H_a \) is \( \mu eq 100 \). The significance level is \( \alpha = 0.05 \).
02

Determine the Test Statistic Formula

For each case, use the formula for the test statistic: \[ z = \frac{\bar{x} - \mu}{\frac{s}{\sqrt{n}}} \] where \( \bar{x} \) is the sample mean, \( \mu \) is the population mean under the null hypothesis, \( s \) is the sample standard deviation, and \( n \) is the sample size.
03

Calculate Test Statistic for Case (a)

Given \( \bar{x} = 103 \), \( s = 11.5 \), and \( n = 65 \), calculate the test statistic: \[ z = \frac{103 - 100}{\frac{11.5}{\sqrt{65}}} \approx 1.827 \]
04

Find the p-value for Case (a)

The test statistic \( z \approx 1.827 \) has a two-tailed p-value, so compute the p-value using the normal distribution. The p-value is approximately 0.068, which is found by finding the area in both tails beyond the test statistic values.
05

Conclusion for Case (a)

Since the p-value \( 0.068 \) is greater than \( \alpha = 0.05 \), we do not reject the null hypothesis. There is insufficient evidence to support \( \mu eq 100 \).
06

Calculate Test Statistic for Case (b)

Given \( \bar{x} = 96.5 \), \( s = 11.0 \), and \( n = 65 \), calculate the test statistic: \[ z = \frac{96.5 - 100}{\frac{11.0}{\sqrt{65}}} \approx -2.726 \]
07

Find the p-value for Case (b)

The test statistic \( z \approx -2.726 \) has a two-tailed p-value. Using the normal distribution, the p-value is approximately 0.007.
08

Conclusion for Case (b)

Since the p-value \( 0.007 \) is less than \( \alpha = 0.05 \), we reject the null hypothesis and conclude there is significant evidence that \( \mu eq 100 \).
09

Calculate Test Statistic for Case (c)

Given \( \bar{x} = 102 \), \( s = 10.5 \), and \( n = 65 \), calculate the test statistic: \[ z = \frac{102 - 100}{\frac{10.5}{\sqrt{65}}} \approx 1.592 \]
10

Find the p-value for Case (c)

The test statistic \( z \approx 1.592 \) corresponds to a two-tailed p-value. The p-value is approximately 0.111 using the normal distribution.
11

Conclusion for Case (c)

Since the p-value \( 0.111 \) is greater than \( \alpha = 0.05 \), we do not reject the null hypothesis. There is insufficient evidence to support \( \mu eq 100 \).

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Key Concepts

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

p-value
The p-value is a crucial concept in hypothesis testing. It helps us decide whether the observed results are statistically significant or happened by random chance. In simpler terms, the p-value tells us the probability of obtaining the test results we did, or more extreme, assuming that the null hypothesis is true.
- A lower p-value indicates that the observed data is unlikely under the null hypothesis, suggesting that the alternative hypothesis might be more plausible.
- In the example above, for each of the cases \(a\), \(b\), and \(c\), the p-value informs us whether the sample mean significantly differs from the hypothesized population mean of 100.

For instance, in Case (b), a p-value of 0.007 indicates a very low probability of obtaining such an extreme test statistic if the null hypothesis were true. This low p-value leads to the rejection of the null hypothesis, suggesting strong evidence against it. Conversely, higher p-values in Cases (a) and (c) suggest insufficient evidence to reject the null hypothesis.
significance level
The significance level, denoted by \(\alpha\), is a threshold set by the researcher to determine whether to reject the null hypothesis. It's essentially a criterion for how extreme the data must be to count as strong enough evidence against the null hypothesis.
- Commonly used significance levels are 0.05, 0.01, and 0.10. In the problem above, the significance level is \(\alpha = 0.05\).
- If the p-value is less than \(\alpha\), we reject the null hypothesis, indicating the results are statistically significant.

Consider Case (b) again, where the p-value of 0.007 is less than our significance level of 0.05. This discrepancy provides a strong basis to reject the null hypothesis, implying that the sample mean is likely different from the population mean of 100. However, in Cases (a) and (c), the p-values 0.068 and 0.111 are both greater than 0.05, so there isn't strong enough evidence to reject the null hypothesis based on our chosen level of significance.
test statistic
The test statistic is a value calculated from the sample data that helps us determine whether to reject the null hypothesis in hypothesis testing. It quantifies the degree of deviation of the sample statistic from the null hypothesis parameter.
- Typically, for means, the test statistic is often a z-score if the population variance is known or based on the sample size and the sample follows normal distribution. The formula used in the provided exercise is:
\[ z = \frac{\bar{x} - \mu}{\frac{s}{\sqrt{n}}} \]
where \(\bar{x}\) is the sample mean, \(\mu\) is the hypothesized population mean, \(s\) is the sample standard deviation, and \(n\) is the sample size.

In Case \(a\), with \(\bar{x} = 103\), \(s = 11.5\), and \(n = 65\), the calculated z-score is approximately 1.827. This value tells us how many standard deviations the sample mean is from the null hypothesis mean.

The test statistic thus serves as an intermediary step for calculating the p-value, which helps us make the decision: reject or not reject the null hypothesis. Understanding the calculation and interpretation of this value is crucial in evaluating the evidence against the null hypothesis.

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Most popular questions from this chapter

The College Board reported that the average number of freshman class applications to public colleges and universities is \(6000(\text {USA Today, December } 26,2002\) ). During a recent application/enrollment period, a sample of 32 colleges and universities showed that the sample mean number of freshman class applications was 5812 with a sample standard deviation of \(1140 .\) Do the data indicate a change in the mean number of applications? Use \(\alpha=.05\).

CCN and ActMedia provided a television channel targeted to individuals waiting in supermarket checkout lines. The channel showed news, short features, and advertisements. The length of the program was based on the assumption that the population mean time a shopper stands in a supermarket checkout line is 8 minutes. A sample of actual waiting times will be used to test this assumption and determine whether actual mean waiting time differs from this standard. a. Formulate the hypotheses for this application. b. \(\quad\) A sample of 120 shoppers showed a sample mean waiting time of 8.5 minutes. Assume a population standard deviation \(\sigma=3.2\) minutes. What is the \(p\) -value? c. \(\quad\) At \(\alpha=.05,\) what is your conclusion? d. Compute a \(95 \%\) confidence interval for the population mean. Does it support your conclusion?

Annual per capita consumption of milk is 21.6 gallons (Statistical Abstract of the United States: 2006 ). Being from the Midwest, you believe milk consumption is higher there and wish to support your opinion. A sample of 16 individuals from the midwestern town of Webster City showed a sample mean annual consumption of 24.1 gallons with a standard deviation of \(s=4.8\) a. Develop a hypothesis test that can be used to determine whether the mean annual consumption in Webster City is higher than the national mean. b. What is a point estimate of the difference between mean annual consumption in Webster City and the national mean? c. \(\quad\) At \(\alpha=.05,\) test for a significant difference. What is your conclusion?

Consider the following hypothesis test: \\[ \begin{array}{l} H_{0}: \mu \leq 50 \\ H_{\mathrm{a}}: \mu>50 \end{array} \\] A sample of 60 is used and the population standard deviation is \(8 .\) Use the critical value approach to state your conclusion for each of the following sample results. Use \\[ \alpha=.05 \\] a. \(\bar{x}=52.5\) b. \(\bar{x}=51\) c. \( \bar{x}=51.8\)

Consider the following hypothesis test: \\[ \begin{array}{l} H_{0}: \mu=15 \\ H_{\mathrm{a}}: \mu \neq 15 \end{array} \\] A sample of 50 provided a sample mean of \(14.15 .\) The population standard deviation is \(3 .\) a. Compute the value of the test statistic. b. What is the \(p\) -value? c. \(\quad\) At \(\alpha=.05,\) what is your conclusion? d. What is the rejection rule using the critical value? What is your conclusion?

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