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Chapter 9: Problem 14

Consider the following hypothesis test: \\[ \begin{array}{l} H_{0}: \mu=22 \\ H_{\mathrm{a}}: \mu \neq 22 \end{array} \\] A sample of 75 is used and the population standard deviation is \(10 .\) Compute the \(p\) -value and state your conclusion for each of the following sample results. Use \(\alpha=.01\) a. \(\bar{x}=23\) b. \(\bar{x}=25.1\) c. \(\bar{x}=20\)

Short Answer

Expert verified
a. Do not reject \(H_0\); b. Reject \(H_0\); c. Do not reject \(H_0\).

Step by step solution

01

Identify the Known Values

The null hypothesis is \( H_0: \mu = 22 \) and the alternative hypothesis is \( H_a: \mu eq 22 \). The sample size is \( n = 75 \) and the population standard deviation is \( \sigma = 10 \). The significance level is \( \alpha = 0.01 \).
02

Calculate the Standard Error

The standard error (SE) is calculated using the formula: \( SE = \frac{\sigma}{\sqrt{n}} \). Given \( \sigma = 10 \) and \( n = 75 \), we find: \[ SE = \frac{10}{\sqrt{75}} \approx 1.1547 \]
03

Standardize the Sample Mean for \( \bar{x} = 23 \)

Calculate the test statistic (z) for \( \bar{x} = 23 \) using the formula: \[ z = \frac{\bar{x} - \mu}{SE} = \frac{23 - 22}{1.1547} \approx 0.866 \]
04

Find the p-value for \( z \) when \( \bar{x} = 23 \)

Since the alternative hypothesis is two-tailed, find the p-value using a standard normal distribution table. For \( z = 0.866 \), the p-value is approximately \( 2 \times (1 - 0.8064) \approx 0.387 \).
05

State the Conclusion for \( \bar{x} = 23 \)

Compare the p-value \( 0.387 \) with \( \alpha = 0.01 \). Since \( 0.387 > 0.01 \), we fail to reject the null hypothesis.
06

Standardize the Sample Mean for \( \bar{x} = 25.1 \)

Calculate the test statistic (z) for \( \bar{x} = 25.1 \): \[ z = \frac{25.1 - 22}{1.1547} \approx 2.686 \]
07

Find the p-value for \( z \) when \( \bar{x} = 25.1 \)

For \( z = 2.686 \), using a standard normal distribution table, the p-value is approximately \( 2 \times (1 - 0.9963) \approx 0.0074 \).
08

State the Conclusion for \( \bar{x} = 25.1 \)

Compare the p-value \( 0.0074 \) with \( \alpha = 0.01 \). Since \( 0.0074 < 0.01 \), we reject the null hypothesis.
09

Standardize the Sample Mean for \( \bar{x} = 20 \)

Calculate the test statistic (z) for \( \bar{x} = 20 \): \[ z = \frac{20 - 22}{1.1547} \approx -1.732 \]
10

Find the p-value for \( z \) when \( \bar{x} = 20 \)

For \( z = -1.732 \), using a standard normal distribution table, the p-value is approximately \( 2 \times (1 - 0.9582) \approx 0.084 \).
11

State the Conclusion for \( \bar{x} = 20 \)

Compare the p-value \( 0.084 \) with \( \alpha = 0.01 \). Since \( 0.084 > 0.01 \), we fail to reject the null hypothesis.

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Key Concepts

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

Null Hypothesis
The null hypothesis is a fundamental concept in hypothesis testing. It represents a default position that indicates no effect or no difference in a study. In mathematical terms, it's expressed as: \( H_0: \mu = \text{value} \). In this particular exercise, the null hypothesis is \( H_0: \mu = 22 \).
This means that we are assuming the population mean is 22 unless we have compelling evidence to suggest otherwise.
Hypothesis testing generally aims to determine whether the observed data deviates enough from what would be expected under the null hypothesis.
  • It essentially acts as a "starting point" for statistical testing.
  • Rejecting or failing to reject the null hypothesis leads to conclusions about the statistical measurements being analyzed.
  • In our case study, failing to reject the null hypothesis implies the sample results are consistent with the population mean being 22.
Alternative Hypothesis
The alternative hypothesis, denoted as \( H_a \), provides an opposing assertion to the null hypothesis. This hypothesis suggests there is a significant effect or difference.
In the given problem, the alternative hypothesis is \( H_a: \mu eq 22 \).
This means we suspect that the population mean is not equal to 22. This could be lower or higher, and we determine this by analyzing our data against certain threshold levels, which are predefined, such as \( \alpha = 0.01 \).
  • The alternative hypothesis is what we may conclude if evidence strongly suggests deviations from the null hypothesis.
  • In the two-tailed test being examined, any sample mean sufficiently different from 22, considering variability, could lead us to reject the null hypothesis.
  • Rejecting the null hypothesis in favor of the alternative indicates statistical evidence that the true mean is not 22.
P-Value
The p-value plays a crucial role in hypothesis testing. It is the probability of obtaining a sample result as extreme as the one observed under the assumption that the null hypothesis is true.
A smaller p-value implies stronger evidence against the null hypothesis.
  • In our scenarios, p-values are compared against the significance level, denoted \( \alpha \).
  • If the p-value is less than \( \alpha \), we reject the null hypothesis.
  • In this exercise, \( \alpha \) is set at 0.01, serving as a strict benchmark.
For example, for a sample mean \( \bar{x} = 25.1 \), the calculated p-value is approximately 0.0074, which is less than 0.01.
This indicates strong evidence against the null hypothesis, leading us to reject it.
Standard Error
The standard error is a measurement that tells us how much the sample mean is expected to fluctuate from the actual population mean.
It is essential in calculating test statistics, such as the z-score.
  • The formula for standard error is given by \( SE = \frac{\sigma}{\sqrt{n}} \), where \( \sigma \) is the population standard deviation and \( n \) is the sample size.
  • In our case, \( \sigma = 10 \) and \( n = 75 \), which results in a standard error of \( SE \approx 1.1547 \).
The standard error allows us to determine how much deviation in the sample mean is due to random chance. This plays a critical role in understanding if an observed deviation in the sample mean is statistically significant or not.
Thus, it helps make robust decisions about rejecting or failing to reject the null hypothesis.

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Most popular questions from this chapter

Because of high production-changeover time and costs, a director of manufacturing must convince management that a proposed manufacturing method reduces costs before the new method can be implemented. The current production method operates with a mean cost of \(\$ 220\) per hour. A research study will measure the cost of the new method over a sample production period. a. Develop the null and alternative hypotheses most appropriate for this study. b. Comment on the conclusion when \(H_{0}\) cannot be rejected. c. Comment on the conclusion when \(H_{0}\) can be rejected.

AOL Time Warner Inc.'s CNN has been the longtime ratings leader of cable television news. Nielsen Media Research indicated that the mean CNN viewing audience was 600,000 viewers per day during 2002 (The Wall Street Journal, March 10,2003 ). Assume that for a sample of 40 days during the first half of \(2003,\) the daily audience was 612,000 viewers with a sample standard deviation of 65,000 viewers. a. What are the hypotheses if CNN management would like information on any change in the CNN viewing audience? b. What is the \(p\) -value? c. Select your own level of significance. What is your conclusion? d. What recommendation would you make to CNN management in this application?

A study by the Centers for Disease Control (CDC) found that \(23.3 \%\) of adults are smokers and that roughly \(70 \%\) of those who do smoke indicate that they want to quit (Associated Press, July 26,2002 ). CDC reported that, of people who smoked at some point in their lives, \(50 \%\) have been able to kick the habit. Part of the study suggested that the success rate for quitting rose by education level. Assume that a sample of 100 college graduates who smoked at some point in their lives showed that 64 had been able to successfully stop smoking. a. State the hypotheses that can be used to determine whether the population of college graduates has a success rate higher than the overall population when it comes to breaking the smoking habit. b. Given the sample data, what is the proportion of college graduates who, having smoked at some point in their lives, were able to stop smoking? c. What is the \(p\) -value? At \(\alpha=.01\), what is your hypothesis testing conclusion?

For the United States, the mean monthly Internet bill is \(\$ 32.79\) per household (CNBC, January 18,2006 ). A sample of 50 households in a southern state showed a sample mean of \(\$ 30.63 .\) Use a population standard deviation of \(\sigma=\$ 5.60\) a. Formulate hypotheses for a test to determine whether the sample data support the conclusion that the mean monthly Internet bill in the southern state is less than the national mean of \(\$ 32.79\) b. What is the value of the test statistic? c. What is the \(p\) -value? d. \(\quad\) At \(\alpha=.01,\) what is your conclusion?

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 \(\alpha=.05\).
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