Step 1: Calculate the test statistic (t)

The formula for the test statistic is:

$t = \frac{M - \mu}{\frac{SD}{\sqrt{n}}}$

Where:

• $M = 70.7$ (sample mean)
• $\mu = 60.8$ (population mean under $H_0$)
• $SD = 17.6$ (sample standard deviation)
• $n = 25$ (sample size)

First, calculate the standard error of the mean (SE): $SE = \frac{SD}{\sqrt{n}} = \frac{17.6}{\sqrt{25}} = \frac{17.6}{5} = 3.52$

Now calculate the test statistic: $t = \frac{M - \mu}{SE} = \frac{70.7 - 60.8}{3.52} = \frac{9.9}{3.52} = 2.813$

So, the test statistic is: $t = 2.813$

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Step 2: Calculate the p-value

Since this is a one-tailed test ($H_a: \mu > 60.8$), the p-value is the area to the right of $t = 2.813$ under a $t$-distribution with $df = n - 1 = 24$.

Using a $t$-distribution table or a calculator: $p = P(T > 2.813 \text{ with } df = 24) = 0.0045$

The p-value is: $p = 0.0045$

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

• $\alpha = 0.005$
• $p = 0.0045$

Since $p < \alpha$, we reject the null hypothesis.

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Step 4: Decision and Conclusion

1. The p-value is less than $\alpha$.
2. This leads to a decision to reject the null hypothesis.
3. The final conclusion is:
   • The sample data support the claim that the population mean is greater than 60.8.

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Summary of Results

• Test statistic: $t = 2.813$
• P-value: $p = 0.0045$
• Decision: Reject the null hypothesis
• Conclusion: The sample data support the claim that the population mean is greater than 60.8.

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Do you need further clarifications or details?

Related Questions

1. How do changes in sample size affect the test statistic and p-value?
2. What if the significance level was 0.01 instead of 0.005—would the decision change?
3. How would the test procedure differ if the standard deviation of the population were known?
4. What is the importance of the degrees of freedom in a t-test?
5. How does this test differ from a two-tailed hypothesis test?

Tip

Always check whether the test is one-tailed or two-tailed, as this directly affects the p-value calculation.