To solve this hypothesis testing problem using the p-value approach, let's go step by step:

Step 1: Identify the given information

• Sample mean ($\bar{x}$) = 30
• Population standard deviation ($\sigma$) = 8
• Sample size ($n$) = 20
• Null hypothesis ($H_0$): $\mu = 25$
• Alternative hypothesis ($H_a$): $\mu > 25$ (right-tailed test)
• Test statistic: $z = 2.795$
• Significance level ($\alpha$) = 0.05

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Step 2: Determine the p-value

For a right-tailed test, the p-value corresponds to the area under the standard normal curve to the right of the test statistic $z = 2.795$.

Using a standard normal table or a calculator:

• Find the cumulative probability for $z = 2.795$: $P(Z \leq 2.795) = 0.9974$
• Therefore, the p-value is: $P(Z > 2.795) = 1 - P(Z \leq 2.795) = 1 - 0.9974 = 0.0026$

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Step 3: Compare the p-value to $\alpha$

• The significance level ($\alpha$) = 0.05.
• The calculated p-value = 0.0026.

Since $\text{p-value} < \alpha$ (0.0026 < 0.05), we reject the null hypothesis.

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Step 4: Conclusion

At the 5% significance level, there is sufficient evidence to conclude that the population mean is greater than 25.

The correct answer is: (a) Reject the null hypothesis. At the 5% significance level, the data provide sufficient evidence that the population mean is greater than 25.

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Relative Questions

1. What is the relationship between the p-value and the significance level in hypothesis testing?
2. How would the conclusion change if the significance level were reduced to 1%?
3. What are the assumptions required for conducting this z-test?
4. How does the test statistic value relate to the rejection region in hypothesis testing?
5. What is the interpretation of a p-value of 0.0026 in practical terms?

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Tip

When performing a hypothesis test, always clearly state whether the test is one-tailed or two-tailed before determining the p-value to avoid errors in your conclusion.