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Chapter 9: Problem 41

Better parking A local high school makes a change that should improve student satisfaction with the parking situation. Before the change, 37% of the school鈥檚 students approved of the parking that was provided. After the change, the principal surveys an SRS of 200 of the over 2500 students at the school. In all, 83 students say that they approve of the new parking arrangement. The principal cites this as evidence that the change was effective. Perform a test of the principal's claim at the \(\alpha=0.05\) significance level.

Short Answer

Expert verified
There is no significant evidence at the 0.05 level that the parking approval rate increased.

Step by step solution

01

Define the Hypotheses

First, we define the null and alternative hypotheses. The null hypothesis ( H_0) is that the proportion of students approving the parking has not changed, so H_0: p = 0.37. The alternative hypothesis ( H_a) is that the proportion has increased, so H_a: p > 0.37.
02

Identify Significance Level 伪

The significance level 伪 is given as 0.05. This means we will reject the null hypothesis if our test statistic reaches a point where the probability of observing such a result under the null hypothesis is less than 5%.
03

Calculate Sample Proportion \( \hat{p} = \frac{x}{n}\)

Now, we'll calculate the sample proportion. \(x = 83\) students approved, and \(n = 200\) students were surveyed. The sample proportion, \(\hat{p} = \frac{83}{200} = 0.415\).
04

Compute Standard Error SE

Calculate the standard error using the formula \[ SE = \sqrt{\frac{p(1-p)}{n}} \]where \(p = 0.37\). Plugging in, \[ SE = \sqrt{\frac{0.37 \times 0.63}{200}} \approx 0.034 \].
05

Compute Test Statistic, \(z\)

We compute the \(z\)-statistic using the formula \[ z = \frac{\hat{p} - p}{SE} \]. Substituting values, \[ z = \frac{0.415 - 0.37}{0.034} \approx 1.324 \].
06

Determine Critical Value and Decision Rule

For a one-tailed test at \(\alpha = 0.05\), the critical \(z\)-value is approximately 1.645. If the computed \(z\)-value is greater than 1.645, we reject the null hypothesis.
07

Conclusion

The \(z\)-statistic of 1.324 is less than the critical value of 1.645. Therefore, we fail to reject the null hypothesis. There is not enough evidence at the \(0.05\) significance level to conclude that the proportion of students who approve of the new parking arrangement has increased.

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Key Concepts

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

Statistical Significance
Statistical significance is a fundamental concept in hypothesis testing. It helps us determine if the observed results are likely due to chance or if they suggest a meaningful effect. For our parking example, we're testing whether the change has genuinely increased student satisfaction. We have set our significance level, denoted by \(\alpha\), at 0.05. Here's what this means:
  • A significance level of 0.05 sets a 5% likelihood that we would mistakenly reject the true null hypothesis, which is the baseline assumption.

  • If the probability of our test statistic under this null hypothesis is less than 5%, we consider it statistically significant, indicating the results are not due to random variation alone.

  • A statistically significant result gives us confidence that the observed change in approval isn't just a fluke.

Understanding and correctly setting the significance level is crucial to making valid statistical inferences.
Null and Alternative Hypotheses
The starting point for any hypothesis test involves two critical assumptions: the null and alternative hypotheses. In our scenario about parking satisfaction:- **Null Hypothesis (\(H_0\))**:
    By default, we assume that there is no change, meaning the proportion of students approving parking remains \(p = 0.37\), the same as before the change.

- **Alternative Hypothesis (\(H_a\))**:
    This represents what we want to support through the test. In this case, \(H_a: p > 0.37\), or that more students approve of the new parking arrangement than before.

It's essential to clearly define these hypotheses because they guide the entire test. The null hypothesis serves as our "no effect" or "no change" benchmark. The specification of \(H_a\) (greater than, less than, or not equal) depends on the research question or claim.
Critical Value
Once we set up our hypotheses, we determine how extreme the data need to be for us to reject the null hypothesis. Enter the critical value, a specific point we calculate based on our chosen significance level and the type of test we're conducting. For the one-tailed test in our parking example at \(\alpha = 0.05\):- **Critical \(z\)-value**: 1.645
    We compare our computed \(z\)-statistic against this critical \(z\)-value.

If the \(z\)-statistic exceeds 1.645, the results are sufficiently extreme; thus, we reject \(H_0\), supporting the alternative hypothesis. In our example, the \(z\)-statistic was calculated to be 1.324, which did not surpass the critical value.
  • This means we don't have enough evidence to refute the null hypothesis, suggesting the parking satisfaction hasn't significantly increased at the 5% significance level.

  • Choosing the correct critical value is essential in marking the boundary for decision making in hypothesis testing.

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Most popular questions from this chapter

Sweetening colas Cola makers test new recipes for loss of sweetness during storage. Trained tasters rate the sweetness before and after storage. From experience, the population distribution of sweetness losses will be close to Normal. Here are the sweetness losses (sweetness before storage minus sweetness after storage) found by tasters from a random sample of 10 batches of a new cola recipe: \(2.0 \quad 0.4 \quad 0.7 \quad 2.0 \quad-0.4 \quad 2.2 \quad-1.3 \quad 1.2 \quad 1.1 \quad 2.3\) Are these data good evidence that the cola lost sweetness? Carry out a test to help you answer this question.

Is it significant? For students without special preparation, SAT Math scores in recent years have varied Normally with mean \(\mu=518\) . One hundred students go through a rigorous training program designed to raise their SAT Math scores by improving their mathematics skills. Use your calculator to carry out a test of $$ \begin{array}{l}{H_{0} : \mu=518} \\ {H_{a} : \mu>518}\end{array} $$ in each of the following situations. (a) The student's scores have mean \(\overline{x}=536.7\) and standard deviation \(s_{x}=114 .\) Is this result significant at the 5\(\%\) level? (b) The students' scores have mean \(\overline{x}=537.0\) and standard deviation \(s_{x}=114 .\) Is this result significant at the 5\(\%\) level? (c) When looked at together, what is the intended lesson of (a) and ( b)?

Check that the conditions for carrying out a one-sample z test for the population proportion p are met. Lefties Simon reads a newspaper report claiming that 12% of all adults in the United States are left- handed. He wonders if 12% of the students at his large public high school are left-handed. Simon chooses an SRS of 100 students and records whether each student is right- or left-handed.

You are testing \(H_{0} : \mu=10\) against \(H_{a} : \mu \neq 10\) based on an SRS of 15 observations from a Normal population. What values of the \(t\) statistic are statistically significant at the \(\alpha=0.005\) level? $$ \begin{array}{ll}{\text { (a) } t>3.326} & {\text { (d) } t<-3.326 \text { or } t>3.326} \\ {\text { (b) } t>3.286} & {\text { (e) } t<-3.286 \text { or } t>3.286}\end{array} $$ (c) \(t > 2.977\)

You use technology to carry out a significance test and get a P-value of 0.031 . The correct conclusion is (a) accept \(H_{a}\) at the \(\alpha=0.05\) significance level. (b) reject \(H_{0}\) at the \(\alpha=0.05\) significance level. (c) reject \(H_{0}\) at the \(\alpha=0.01\) significance level. (d) fail to reject \(H_{0}\) at the \(\alpha=0.05\) significance level. (e) fail to reject \(H_{a}\) at the \(\alpha=0.05\) significance level.
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