Questions: Complete the hypothesis test with the alternative hypothesis μd>0 based on the paired data that follow and d=B-A. Use α=0.05. Assume normality. A 700 830 860 1080 930 B 720 820 890 1100 960 B: Solve using the evalue approach. n= , 𝑑̅=20 x, 5 d=10 Hypothesis Test: H0: μd=0 vs. Ha: μd>0 Solve for t. t =4.47 Use t to find the p-value. p-value: 0.0025 Decision for the p-value Approach: Reject H0, p-value < α Find t(df, a). t(df, a)=2.353 Decision for the Classical Approach: Reject H0, t >t(df, a)

Complete the hypothesis test with the alternative hypothesis μd>0 based on the paired data that follow and d=B-A. Use α=0.05. Assume normality.

A 700 830 860 1080 930 B 720 820 890 1100 960

B: Solve using the evalue approach. n= , 𝑑̅=20 x, 5 d=10

Hypothesis Test:
H0: μd=0 vs. Ha: μd>0

Solve for t.
t =4.47

Use t to find the p-value. p-value: 0.0025

Decision for the p-value Approach: Reject H0, p-value < α

Find t(df, a).
t(df, a)=2.353

Decision for the Classical Approach: Reject H0, t >t(df, a)

Transcript text: Complete the hypothesis test with the alternative hypothesis $\mu_{d}>0$ based on the paired data that follow and $d=B-A$. Use $\alpha=0.05$. Assume normality. A 7008308601080930 B 7208208901100960 B: Solve using the evalue approach. $n=$ $\qquad$ , $\bar{d}=20$ $x, 5 d=10$ Hypothesis Test: \[ \begin{array}{l} H_{0}: \mu_{d}=0 \text { vs. } H_{a}: \mu_{d}>0 \\ \end{array} \] Solve for $t$. \[ t *=4.47 \] Use t* to find the p-value. p -value: $0.0025$ Decision for the $p$-value Approach: Reject $\mathrm{H}_{0}$, p-value < $\alpha$ Find $t(d f, a)$. \[ t(d f, a)=2.353 \] Decision for the Classical Approach: Reject $H_{0}, t *>t(d f, a)$

Solution

Solution Steps

Step 1: Data Summary

We have paired data from two groups, $ A $ and $ B $, with the following values:

$ A = [700, 830, 860, 1080, 930] $
$ B = [720, 820, 890, 1100, 960] $

The differences $ d = B - A $ are calculated as follows: \[ d = [20, -10, 30, 20, 30] \]

Step 2: Calculate Sample Statistics

The sample size $ n $ is: \[ n = 5 \]

The mean of the differences $ \bar{d} $ is: \[ \bar{d} = \frac{20 + (-10) + 30 + 20 + 30}{5} = 18.0 \]

The sample standard deviation $ s_d $ is calculated as: \[ s_d = \sqrt{\frac{\sum (d_i - \bar{d})^2}{n - 1}} \approx 16.4317 \]

The standard error (SE) of the mean difference is: \[ SE = \frac{s_d}{\sqrt{n}} = \frac{16.4317}{\sqrt{5}} \approx 7.3485 \]

Step 3: Calculate Test Statistic

The test statistic $ t $ is calculated using the formula: \[ t = \frac{\bar{d}}{SE} = \frac{18.0}{7.3485} \approx 2.4489 \]

Step 4: Determine Critical Value and p-value

For a two-tailed test at $ \alpha = 0.05 $ with $ df = n - 1 = 4 $, the critical value $ t_{\alpha/2, df} $ is: \[ t_{(0.025, 4)} \approx 2.7764 \]

The p-value is calculated as: \[ P = 2 \times (1 - T(|t|)) = 2 \times (1 - T(2.4489)) \approx 0.0705 \]

Step 5: Decision Making

Since $ p \text{-value} = 0.0705 $ is greater than $ \alpha = 0.05 $, we fail to reject the null hypothesis $ H_0 $.
The test statistic $ t \approx 2.4489 $ is less than the critical value $ t_{(0.025, 4)} \approx 2.7764 $, confirming the decision to fail to reject $ H_0 $.

Final Answer

The conclusion of the hypothesis test is that we fail to reject the null hypothesis $ H_0: \mu_d = 0 $ in favor of the alternative hypothesis $ H_a: \mu_d > 0 $.

$\boxed{\text{Fail to reject } H_0}$

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