A university administrator claims that no more than $55 \%$ of students at their school own a car. A random sample of 80 students found that 48 own a car. Test this claim at a significance level of 0.05.

Claim: p ? ? - $H _{\text {? }}: p$ ? ?
Opposite: p ? ? - $H _?: p$ ? ?
Test Statistic:
Critical Value:
Reject $H _0$ or Fail to Reject $H _0$
Does the data appear to support the claim?

Question

02-11-2026
Mathematics

Answered

A university administrator claims that no more than $55 \%$ of students at their school own a car. A random sample of 80 students found that 48 own a car. Test this claim at a significance level of 0.05.

Claim: p ? ? - $H _{\text {? }}: p$ ? ?
Opposite: p ? ? - $H _?: p$ ? ?
Test Statistic:
Critical Value:
Reject $H _0$ or Fail to Reject $H _0$
Does the data appear to support the claim?

Verified Answer :

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GinnyAnswer GinnyAnswer · Answer 1 · 2026-11-02T00:08:38-05:00

Claim: $p \le 0.55$, Null Hypothesis ($H_0$): $p \le 0.55$, Alternative Hypothesis ($H_a$): $p > 0.55$.
Test Statistic: $Z = \frac{0.6 - 0.55}{\sqrt{\frac{0.55 \times 0.45}{80}}} \approx 0.899$.
Critical Value: $Z_{0.05} \approx 1.645$.
Decision: Since $0.899 < 1.645$, Fail to reject $H_0$. The data supports the claim that the proportion is no more than 55%.
Final Answer: $\boxed{\text{Fail to reject } H_0}$

Explanation

Setup and Hypotheses

The university administrator claims that the population proportion of students who own a car is no more than 55%. We are testing this claim using a sample of 80 students where 48 own a car.

Claim and Hypotheses:

The claim is that the proportion $p$ is no more than 0.55.

Claim: $p \le 0.55$.
Null Hypothesis ($H_0$): $p \le 0.55$.
Alternative Hypothesis ($H_a$): $p > 0.55$.

This is a right-tailed test because we are testing for a proportion greater than the specified value.

Sample Data:

Sample size $n = 80$.
Number of students owning a car $x = 48$.
Sample proportion $\hat{p} = \frac{x}{n} = \frac{48}{80} = 0.6$.

Calculate the Test Statistic

To determine if our sample results significantly differ from the administrator's claim, we calculate the test statistic using the formula for a one-proportion z-test:
\[Z = \frac{\hat{p} - p_0}{\sqrt{\frac{p_0(1 - p_0)}{n}}}\]
where $p_0 = 0.55$ and $n = 80$.

Standard Error: $\sqrt{\frac{0.55 \times (1 - 0.55)}{80}} = \sqrt{\frac{0.55 \times 0.45}{80}} = \sqrt{\frac{0.2475}{80}} \approx 0.0556$.
Test Statistic: $Z = \frac{0.6 - 0.55}{0.0556} \approx 0.8989$.

Determine Critical Value and Conclusion

For a significance level $\alpha = 0.05$ in a right-tailed test, we look for the critical z-value that separates the top 5% of the standard normal distribution.

Using the standard normal distribution, the critical value $Z_{0.05} \approx 1.645$.

Now, compare our calculated test statistic $Z = 0.8989$ with the critical value $Z_{critical} = 1.645$:

Since $0.8989 < 1.645$, our test statistic does not fall into the rejection region.

Conclusion: We fail to reject the null hypothesis $H_0$. There is not sufficient evidence to conclude that the proportion of students who own a car is greater than 55%. Therefore, the data appears to support the administrator's claim.

Examples

In real life, this method is used by researchers and businesses to validate claims against data. For example, a quality control manager might claim that no more than 5% of manufactured parts are defective. By testing a random sample, they can statistically determine if the production process meets that standard or if there is a problem that needs investigation, using a formula like $Z = \frac{\hat{p} - p_0}{\sqrt{\frac{p_0(1-p_0)}{n}}}$ to decide if the observed defect rate is significantly high.