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Chapter 9: Problem 12

Consider the following hypothesis test: \\[ \begin{array}{l} H_{0}: \mu \geq 80 \\ H_{\mathrm{a}}: \mu<80 \end{array} \\] A sample of 100 is used and the population standard deviation is \(12 .\) Compute the \(p\) -value and state your conclusion for each of the following sample results. Use \(\alpha=.01\) a. \(\quad \bar{x}=78.5\) b. \(\bar{x}=77\) c. \(\quad \bar{x}=75.5\) d. \(\bar{x}=81\)

Short Answer

Expert verified
Reject \(H_0\) for b and c; do not reject \(H_0\) for a and d.

Step by step solution

01

Calculate the Z-Score

For each sample mean, we start by calculating the Z-score using the formula:\[ Z = \frac{\bar{x} - \mu_0}{\frac{\sigma}{\sqrt{n}}} \]where \(\bar{x}\) is the sample mean, \(\mu_0 = 80\) is the hypothesized population mean, \(\sigma = 12\) is the population standard deviation, and \(n = 100\) is the sample size.For part (a), \(\bar{x} = 78.5\):\[ Z = \frac{78.5 - 80}{\frac{12}{\sqrt{100}}} = \frac{-1.5}{1.2} = -1.25 \]
02

Calculate the p-Value for Part (a)

We next find the p-value for the calculated Z-score. For a Z-score of -1.25, we use a Z-table or a calculator: - The p-value is the probability that a standard normal variable is less than -1.25, which is approximately 0.1056.
03

Compare p-Value to Significance Level for Part (a)

The significance level \(\alpha\) is 0.01. We compare the p-value to \(\alpha\):- Since 0.1056 > 0.01, we do not reject the null hypothesis \(H_0\) for part (a).
04

Repeat Z-Score Calculation for Part (b)

For part (b), \(\bar{x} = 77\):\[ Z = \frac{77 - 80}{\frac{12}{\sqrt{100}}} = \frac{-3}{1.2} = -2.5 \]
05

Calculate the p-Value for Part (b)

Finding the p-value for a Z-score of -2.5: - The p-value is approximately 0.0062.
06

Compare p-Value to Significance Level for Part (b)

Comparing the p-value to \(\alpha\):- Since 0.0062 < 0.01, we reject the null hypothesis \(H_0\) for part (b).
07

Repeat Z-Score Calculation for Part (c)

For part (c), \(\bar{x} = 75.5\):\[ Z = \frac{75.5 - 80}{\frac{12}{\sqrt{100}}} = \frac{-4.5}{1.2} = -3.75 \]
08

Calculate the p-Value for Part (c)

Finding the p-value for a Z-score of -3.75: - The p-value is approximately 0.0001.
09

Compare p-Value to Significance Level for Part (c)

Comparing the p-value to \(\alpha\):- Since 0.0001 < 0.01, we reject the null hypothesis \(H_0\) for part (c).
10

Repeat Z-Score Calculation for Part (d)

For part (d), \(\bar{x} = 81\):\[ Z = \frac{81 - 80}{\frac{12}{\sqrt{100}}} = \frac{1}{1.2} = 0.83 \]
11

Calculate the p-Value for Part (d)

Finding the p-value for a Z-score of 0.83:- The p-value is the probability that a standard normal variable is less than 0.83, which is about 0.7967. However, since this is a one-tailed test to the left, we actually need to subtract from 1, which gives \(1 - 0.7967 = 0.2033\).
12

Compare p-Value to Significance Level for Part (d)

Comparing the p-value to \(\alpha\):- Since 0.7967 > 0.01, we do not reject the null hypothesis \(H_0\) for part (d).

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Key Concepts

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

Z-Score Calculation
Understanding the z-score is essential in hypothesis testing. The z-score helps convert your sample mean into a form that can be compared against a standard normal distribution. This allows us to assess how unusual our sample results are under the null hypothesis. The formula for calculating the z-score is \[Z = \frac{\bar{x} - \mu_0}{\frac{\sigma}{\sqrt{n}}}\]where:
  • \( \bar{x} \) is the sample mean,
  • \( \mu_0 \) is the hypothesized population mean,
  • \( \sigma \) is the population standard deviation,
  • \( n \) is the sample size.
By plugging in these values, you convert your sample mean to see how many standard deviations away it is from the population mean.
A higher absolute z-score indicates a sample mean further from the null hypothesis, indicating that our observed data might be less likely if the null hypothesis is true.
P-Value Interpretation
The p-value is a way to measure the strength of evidence against the null hypothesis. Once you calculate the z-score, you can find the corresponding p-value using a Z-table or statistical software.
  • A small p-value (usually \(< 0.05\) or \(< 0.01\), depending on the chosen significance level) indicates strong evidence against the null hypothesis.
  • A large p-value suggests that the null hypothesis cannot be rejected, indicating the results could easily occur if the null hypothesis is true.
P-values help quantify the evidence against the null hypothesis but do not prove it's false. In decision-making, comparing the p-value to the predefined significance level \(\alpha\) is crucial.
Significance Level
The significance level, represented as \(\alpha\), is the threshold for deciding whether to reject the null hypothesis. It is set before the test is conducted and often chosen as 0.01, 0.05, or 0.10, representing a 1%, 5%, or 10% risk of rejecting the null hypothesis when it is actually true.
  • If the p-value is less than \(\alpha\), there is sufficient evidence to reject the null hypothesis. This implies that the sample provides strong enough data to suggest the alternative hypothesis may be true.
  • If the p-value is greater than \(\alpha\), the null hypothesis is not rejected, implying insufficient evidence to support the alternative hypothesis.
Choosing a proper significance level is essential since it reflects how much risk of error you are willing to accept. A smaller \(\alpha\) lowers the risk, leading to stricter criteria for rejecting the null hypothesis and potentially requiring more substantial evidence.

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Most popular questions from this chapter

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?

At Western University the historical mean of scholarship examination scores for freshman applications is \(900 .\) A historical population standard deviation \(\sigma=180\) is assumed known. Each year, the assistant dean uses a sample of applications to determine whether the mean examination score for the new freshman applications has changed. a. State the hypotheses. b. What is the \(95 \%\) confidence interval estimate of the population mean examination score if a sample of 200 applications provided a sample mean of \(\bar{x}=935 ?\) c. Use the confidence interval to conduct a hypothesis test. Using \(\alpha=.05,\) what is your conclusion? d. What is the \(p\) -value?

The chamber of commerce of a Florida Gulf Coast community advertises that area residential property is available at a mean cost of \(\$ 125,000\) or less per lot. Suppose a sample of 32 properties provided a sample mean of \(\$ 130,000\) per lot and a sample standard deviation of \(\$ 12,500 .\) Use a .05 level of significance to test the validity of the advertising claim.

The National Center for Health Statistics released a report that stated \(70 \%\) of adults do not exercise regularly (Associated Press, April 7, 2002). A researcher decided to conduct a study to see whether the claim made by the National Center for Health Statistics differed on a state-by-state basis. a. State the null and alternative hypotheses assuming the intent of the researcher is to identify states that differ from the \(70 \%\) reported by the National Center for Health Statistics. b. \(\quad\) At \(\alpha=.05,\) what is the research conclusion for the following states: Wisconsin: 252 of 350 adults did not exercise regularly California: 189 of 300 adults did not exercise regularly

Consider the following hypothesis test: \\[ \begin{array}{l} H_{0}: \mu \geq 20 \\ H_{\mathrm{a}}: \mu<20 \end{array} \\] A sample of 50 provided a sample mean of \(19.4 .\) The population standard deviation is \(2 .\) a. Compute the value of the test statistic. b. What is the \(p\) -value? c. Using \(\alpha=.05,\) what is your conclusion? d. What is the rejection rule using the critical value? What is your conclusion?
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