Questions: Assume that the differences are normally distributed. Complete parts (a) through (d). Observation 1 2 3 4 5 6 7 8 Xi 44.9 48.1 45.3 39.7 46.4 53.8 45.5 43.3 Yi 45.1 48.5 47.8 45.1 48.2 53.9 48.4 45.5 sd=1.775 (Round to three decimal places as needed.) (c) Test if μd<0 at the α=0.05 level of significance. What are the correct null and alternative hypotheses? A. H0: μd=0 B. H0: μd>0 H1: μd<0 H1: μd<0 C. H0: μd<0 D. H0: μd<0 H1: μd>0 H1: μd=0 Find the test statistic for this hypothesis test. t0=-3.09 (Round to two decimal places as needed.) Determine the P -value for this hypothesis test. P -value = 0.010 (Round to three decimal places as needed.) Choose the correct conclusion. A. Do not reject the null hypothesis. There is sufficient evidence that μd<0 at the α=0.05 level of significance. B. Reject the null hypothesis. There is sufficient evidence that μd<0 at the α=0.05 level of significance. C. Do not reject the null hypothesis. There is insufficient evidence that μd<0 at the α=0.05 level of significance. D. Reject the null hypothesis. There is insufficient evidence that μd<0 at the α=0.05 level of significance. (d) Compute a 95% confidence interval about the population mean difference μd. The lower bound is The upper bound is (Round to two decimal places as needed.)

Assume that the differences are normally distributed. Complete parts (a) through (d).

Observation 1 2 3 4 5 6 7 8
Xi 44.9 48.1 45.3 39.7 46.4 53.8 45.5 43.3
Yi 45.1 48.5 47.8 45.1 48.2 53.9 48.4 45.5
sd=1.775 (Round to three decimal places as needed.)

(c) Test if μd<0 at the α=0.05 level of significance.

What are the correct null and alternative hypotheses?
A. H0: μd=0 B. H0: μd>0 H1: μd<0 H1: μd<0
C. H0: μd<0 D. H0: μd<0
H1: μd>0
H1: μd=0

Find the test statistic for this hypothesis test.
t0=-3.09 (Round to two decimal places as needed.)

Determine the P -value for this hypothesis test.
P -value = 0.010 (Round to three decimal places as needed.)

Choose the correct conclusion.
A. Do not reject the null hypothesis. There is sufficient evidence that μd<0 at the α=0.05 level of significance.
B. Reject the null hypothesis. There is sufficient evidence that μd<0 at the α=0.05 level of significance.
C. Do not reject the null hypothesis. There is insufficient evidence that μd<0 at the α=0.05 level of significance.
D. Reject the null hypothesis. There is insufficient evidence that μd<0 at the α=0.05 level of significance.

(d) Compute a 95% confidence interval about the population mean difference μd.

The lower bound is 
The upper bound is 
(Round to two decimal places as needed.)
Transcript text: Assume that the differences are normally distributed. Complete parts (a) through (d). \begin{tabular}{lcccccccc} Observation & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 \\ $X_{i}$ & 44.9 & 48.1 & 45.3 & 39.7 & 46.4 & 53.8 & 45.5 & 43.3 \\ $Y_{i}$ & 45.1 & 48.5 & 47.8 & 45.1 & 48.2 & 53.9 & 48.4 & 45.5 \end{tabular} $s_{d}=1.775$ (Round to three decimal places as needed.) (c) Test if $\mu_{d}<0$ at the $\alpha=0.05$ level of significance. What are the correct null and alternative hypotheses? A. $H_{0}: \mu_{d}=0$ B. $H_{0}: \mu_{d}>0$ $H_{1}: \mu_{d}<0$ $H_{1}: \mu_{d}<0$ C. $H_{0}: \mu_{d}<0$ D. $H_{0}: \mu_{d}<0$ \[ H_{1}: \mu_{d}>0 \] \[ \mathrm{H}_{1}: \mu_{\mathrm{d}}=0 \] Find the test statistic for this hypothesis test. \[ t_{0}=-3.09 \text { (Round to two decimal places as needed.) } \] Determine the P -value for this hypothesis test. P -value $=$ $\square$ 0.010 (Round to three decimal places as needed.) Choose the correct conclusion. A. Do not reject the null hypothesis. There is sufficient evidence that $\mu_{\mathrm{d}}<0$ at the $\alpha=0.05$ level of significance. B. Reject the null hypothesis. There is sufficient evidence that $\mu_{\mathrm{d}}<0$ at the $\alpha=0.05$ level of significance. C. Do not reject the null hypothesis. There is insufficient evidence that $\mu_{\mathrm{d}}<0$ at the $\alpha=0.05$ level of significance. D. Reject the null hypothesis. There is insufficient evidence that $\mu_{\mathrm{d}}<0$ at the $\alpha=0.05$ level of significance. (d) Compute a $95 \%$ confidence interval about the population mean difference $\mu_{\mathrm{d}}$. The lower bound is $\square$ The upper bound is $\square$ (Round to two decimal places as needed.)
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Solution

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Solution Steps

Step 1: Calculate the Differences

First, we calculate the differences di=XiYi d_i = X_i - Y_i for each observation: d=[0.2,0.4,2.5,5.4,1.8,0.1,2.9,2.2] d = [-0.2, -0.4, -2.5, -5.4, -1.8, -0.1, -2.9, -2.2]

Step 2: Compute the Mean Difference

Next, we compute the mean of the differences: dˉ=din=1.93751=1.9375 \bar{d} = \frac{\sum d_i}{n} = \frac{-1.9375}{1} = -1.9375

Step 3: Calculate the Test Statistic

We calculate the test statistic t0 t_0 using the formula: t0=dˉ0sd/n=1.937501.775/83.09 t_0 = \frac{\bar{d} - 0}{s_d / \sqrt{n}} = \frac{-1.9375 - 0}{1.775 / \sqrt{8}} \approx -3.09

Step 4: Determine the P-value

The P-value for the test is calculated using the cumulative distribution function of the t-distribution: P-value=P(T<t0)0.0088 P\text{-value} = P(T < t_0) \approx 0.0088

Step 5: Make a Conclusion

We compare the P-value to the significance level α=0.05 \alpha = 0.05 : Since 0.0088<0.05 0.0088 < 0.05 , we reject the null hypothesis H0:μd=0 H_0: \mu_d = 0 . There is sufficient evidence that μd<0 \mu_d < 0 .

Step 6: Compute the 95% Confidence Interval

We calculate the 95% confidence interval for the population mean difference μd \mu_d : Margin of Error=tcriticalsdn1.18896 \text{Margin of Error} = t_{\text{critical}} \cdot \frac{s_d}{\sqrt{n}} \approx 1.18896 The confidence interval is given by: (dˉMargin of Error,dˉ+Margin of Error)=(3.1265,0.7485) \left( \bar{d} - \text{Margin of Error}, \bar{d} + \text{Margin of Error} \right) = \left(-3.1265, -0.7485\right)

Final Answer

The correct null and alternative hypotheses are A. H0:μd=0 H_{0}: \mu_{d}=0 and H1:μd<0 H_{1}: \mu_{d}<0 .
The test statistic is t0=3.09 t_{0} = -3.09 .
The P-value is 0.009 0.009 .
The correct conclusion is B. Reject the null hypothesis. There is sufficient evidence that μd<0 \mu_{d}<0 at the α=0.05 \alpha=0.05 level of significance.
The lower bound of the 95% 95\% confidence interval is 3.13 -3.13 and the upper bound is 0.75 -0.75 .
Thus, the confidence interval is (3.13,0.75) (-3.13, -0.75) .

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