To test the claim about the mean of the differences $\mu_d$, we perform a one-sample $t$-test for the mean of paired differences.

Given:

• Null hypothesis: $H_0: \mu_d \geq 0$
• Alternative hypothesis: $H_a: \mu_d < 0$ (one-tailed test)
• $\alpha = 0.05$
• Sample mean difference: $\bar{d} = 1.5$
• Standard deviation of differences: $s_d = 3.3$
• Sample size: $n = 17$

Test statistic formula:

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

• $\bar{d}$ = sample mean difference
• $\mu_0$ = hypothesized mean difference (0 in this case)
• $s_d$ = sample standard deviation of differences
• $n$ = sample size

Calculate the test statistic:

$t = \frac{1.5 - 0}{3.3 / \sqrt{17}} = \frac{1.5}{3.3 / 4.123} = \frac{1.5}{0.8007} \approx 1.872$

Degrees of freedom:

$df = n - 1 = 17 - 1 = 16$

Critical value:

Using a one-tailed $t$-distribution table at $\alpha = 0.05$ and $df = 16$, the critical $t$-value is approximately: $t_{critical} = -1.746$

Decision rule:

• If $t \leq t_{critical}$, reject $H_0$.
• If $t > t_{critical}$, fail to reject $H_0$.

Compare $t$:

The calculated $t = 1.872$ is greater than $t_{critical} = -1.746$.

Conclusion:

We fail to reject $H_0$. There is not enough evidence to support the claim that $\mu_d < 0$ at $\alpha = 0.05$.

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Would you like to see details on how to interpret the conclusion or any related topics?

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Related Questions:

1. What does it mean to fail to reject the null hypothesis in this context?
2. How do changes in sample size affect the $t$-test?
3. What is the role of $\alpha$ in hypothesis testing?
4. Can this test be performed using software or calculators? If so, how?
5. How does assuming normality of differences affect the validity of this test?

Tip: Always verify the assumptions of normality when performing $t$-tests for small sample sizes.