To test the claim about the mean of the differences ($\mu_d$) for a population of paired data, we follow the steps of hypothesis testing for a one-sample $t$-test for means.

Step 1: Define the hypotheses

• Null hypothesis ($H_0$): $\mu_d = 0$
• Alternative hypothesis ($H_a$): $\mu_d < 0$ (claim)

This is a one-tailed test.

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Step 2: Test statistic formula

The test statistic for a $t$-test is:

$t = \frac{\bar{d} - \mu_0}{\frac{s_d}{\sqrt{n}}}$

Where:

• $\bar{d}$ = sample mean of differences = 1.3
• $\mu_0$ = hypothesized mean difference = 0
• $s_d$ = standard deviation of differences = 3.6
• $n$ = sample size = 18

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Step 3: Calculate the test statistic

$t = \frac{1.3 - 0}{\frac{3.6}{\sqrt{18}}}$

1. Calculate the denominator: $\text{Standard error} = \frac{3.6}{\sqrt{18}} \approx \frac{3.6}{4.24} \approx 0.849$

2. Calculate $t$: $t = \frac{1.3}{0.849} \approx 1.53$

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Step 4: Determine the critical value

The critical value for a one-tailed $t$-test at $\alpha = 0.05$ with $df = n - 1 = 17$ can be obtained from a $t$-distribution table or using software.

For $t_{0.05, 17}$: $t_{\text{critical}} \approx -1.740$

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Step 5: Decision rule

• If $t$-statistic $< t_{\text{critical}}$, reject $H_0$.
• Otherwise, fail to reject $H_0$.

Here: $t = 1.53 \quad \text{and} \quad t_{\text{critical}} = -1.740$

Since $t > t_{\text{critical}}$, we fail to reject the null hypothesis.

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Step 6: Conclusion

There is not enough evidence at the $\alpha = 0.05$ level of significance to support the claim that $\mu_d < 0$.

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Questions to Consider:

1. What is the importance of checking whether the populations are normally distributed in paired $t$-tests?
2. How would the conclusion change if the sample size were larger or smaller?
3. Why is the alternative hypothesis formulated as $\mu_d < 0$?
4. How do we interpret the p-value in this context if it were calculated?
5. What would the decision be if $\alpha$ were 0.10 instead of 0.05?

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Tip:

When performing hypothesis tests, always visualize the data if possible to ensure assumptions (such as normality) are not violated.