Questions: Test the claim about the population mean μ at the level of significance α. Assume the population is normally distributed. Claim: μ<4915 ; α=0.05 Sample statistics: x̄=5017, s=5613, n=56 What are the null and alternative hypotheses? H₀: μ=4915 Hₐ: μ<4915 (Type integers or decimals. Do not round.)

Test the claim about the population mean μ at the level of significance α. Assume the population is normally distributed. Claim: μ<4915 ; α=0.05 Sample statistics: x̄=5017, s=5613, n=56 What are the null and alternative hypotheses? H₀: μ=4915 Hₐ: μ<4915 (Type integers or decimals. Do not round.)
Transcript text: Test the claim about the population mean $\mu$ at the level of significance $\alpha$. Assume the population is normally distributed. Claim: $\mu<4915 ; \alpha=0.05$ Sample statistics: $\bar{x}=5017, s=5613, n=56$ What are the null and alternative hypotheses? $\mathrm{H}_{0}$ : $\square$ $\square$ $\square$ $\mathrm{H}_{\mathrm{a}}$ : $\square$ $\square$ $\square$ (Type integers or decimals. Do not round.)
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Solution

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Solution Steps

Step 1: State the Hypotheses

We are testing the claim about the population mean \( \mu \). The null and alternative hypotheses are defined as follows:

\[ H_0: \mu = 4915 \] \[ H_a: \mu < 4915 \]

Step 2: Calculate the Standard Error

The standard error \( SE \) is calculated using the formula:

\[ SE = \frac{\sigma}{\sqrt{n}} = \frac{5613}{\sqrt{56}} \approx 750.0687 \]

Step 3: Calculate the Test Statistic

The test statistic \( Z_{test} \) is calculated using the formula:

\[ Z_{test} = \frac{\bar{x} - \mu_0}{SE} = \frac{5017 - 4915}{750.0687} \approx 0.136 \]

Step 4: Calculate the P-value

For a left-tailed test, the P-value is determined as follows:

\[ P = T(z) \approx 0.5541 \]

Step 5: Conclusion

At a significance level of \( \alpha = 0.05 \), we compare the P-value with \( \alpha \):

\[ P \approx 0.5541 > 0.05 \]

Since the P-value is greater than the significance level, we fail to reject the null hypothesis \( H_0 \).

Final Answer

The conclusion is that there is not enough evidence to support the claim that the population mean \( \mu \) is less than 4915. Thus, we box the final answer:

\[ \boxed{H_0 \text{ is not rejected}} \]

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