The following sample of eight matched pairs

Question

sample 1: 18, 25, 12, 19, 15, 16, 23, 24
sample 2: 23, 21, 11, 18, 19, 10, 14, 17
Let uv1 and uv2 represent the population means and let uvd = uv1-uv2.
A test will be made of the hypothesis Hv0 : uvd = 0 versus Hv1 : uvd > 0.

Compute the differences.

Answer 1

To compute the differences for each paired sample, subtract each value in Sample 2 from the corresponding value in Sample 1. Here are the pairs and their differences:

1. \(18 - 23 = -5\)
2. \(25 - 21 = 4\)
3. \(12 - 11 = 1\)
4. \(19 - 18 = 1\)
5. \(15 - 19 = -4\)
6. \(16 - 10 = 6\)
7. \(23 - 14 = 9\)
8. \(24 - 17 = 7\)

Now we can summarize the differences:

• Differences: \(-5, 4, 1, 1, -4, 6, 9, 7\)

If we want to find the mean of these differences, we can proceed by calculating the average:

\[ \text{Mean of differences} = \frac{\sum \text{differences}}{n} \]

Where \(n = 8\) (the number of pairs).

Calculating the sum of the differences:

\[ -5 + 4 + 1 + 1 - 4 + 6 + 9 + 7 = 19 \]

Now, dividing by \(n\):

\[ \text{Mean of differences} = \frac{19}{8} = 2.375 \]

Thus, the differences array is:

• Differences: \(-5, 4, 1, 1, -4, 6, 9, 7\)
• Mean of differences: \(2.375\)

Answer 2

(b) Compute the test statistic. Round your final answer to at least three decimal places.

Answer 3

To compute the test statistic for the hypothesis test \(H_{0}: \mu_d = 0\) versus \(H_{1}: \mu_d > 0\), we will use the formula for the t-test for paired samples:

\[ t = \frac{\bar{d}}{s_d / \sqrt{n}} \]

Where:

• \(\bar{d}\) is the mean of the differences.
• \(s_d\) is the standard deviation of the differences.
• \(n\) is the number of pairs (which is 8 in this case).

Step 1: Calculate the Mean of the Differences (\(\bar{d}\))

As calculated previously:

\[ \bar{d} = 2.375 \]

Step 2: Calculate the Standard Deviation of the Differences (\(s_d\))

1. First, calculate the squared differences from the mean:

\[ \text{Differences (d)}: -5, 4, 1, 1, -4, 6, 9, 7 \]

Mean (\(\bar{d}\)) = 2.375.

Now, calculate each squared difference:

\[ (-5 - 2.375)^2 = (-7.375)^2 = 54.390625 \ (4 - 2.375)^2 = (1.625)^2 = 2.640625 \ (1 - 2.375)^2 = (-1.375)^2 = 1.890625 \ (1 - 2.375)^2 = (-1.375)^2 = 1.890625 \ (-4 - 2.375)^2 = (-6.375)^2 = 40.590625 \ (6 - 2.375)^2 = (3.625)^2 = 13.140625 \ (9 - 2.375)^2 = (6.625)^2 = 43.890625 \ (7 - 2.375)^2 = (4.625)^2 = 21.390625 \]

2. Now, sum up the squared differences:

\[ \text{Sum of squared differences} = 54.390625 + 2.640625 + 1.890625 + 1.890625 + 40.590625 + 13.140625 + 43.890625 + 21.390625 = 139.875 \]

3. Calculate the variance and then the standard deviation:

\[ s_d^2 = \frac{\text{Sum of squared differences}}{n - 1} = \frac{139.875}{8 - 1} = \frac{139.875}{7} \approx 19.981 \]

\[ s_d \approx \sqrt{19.981} \approx 4.472 \]

Step 3: Calculate the test statistic \(t\)

Now we can compute the test statistic:

\[ t = \frac{\bar{d}}{s_d / \sqrt{n}} = \frac{2.375}{4.472/\sqrt{8}} = \frac{2.375}{4.472/2.828} = \frac{2.375}{1.58} \approx 1.501 \]

Final Answer

Thus, the test statistic \(t\) is:

\[ \boxed{1.501} \]

Make sure to round your final answer to at least three decimal places as required.

Answer 4

Can you reject Hv0 at the a-0.05 level

Answer 5

To determine whether to reject the null hypothesis \(H_{0}: \mu_d = 0\) at the \(\alpha = 0.05\) significance level, we need to compare the calculated test statistic with the critical value from the t-distribution.

Step 1: Determine the Degrees of Freedom

The degrees of freedom (\(df\)) for a paired sample t-test is given by:

\[ df = n - 1 \]

where \(n\) is the number of pairs. In this case, \(n = 8\):

\[ df = 8 - 1 = 7 \]

Step 2: Find the Critical t-value

Using a t-table or a calculator for the one-tailed test at \(\alpha = 0.05\) with \(df = 7\):

The critical t-value for a one-tailed test with \(\alpha = 0.05\) and 7 degrees of freedom is approximately:

\[ t_{critical} \approx 1.895 \]

Step 3: Compare the Test Statistic with the Critical Value

We previously calculated the test statistic as:

\[ t \approx 1.501 \]

Now, we compare it with the critical value:

• Test Statistic: \(t \approx 1.501\)
• Critical Value: \(t_{critical} \approx 1.895\)

Decision Rule

• If \(t > t_{critical}\), we reject \(H_{0}\).
• If \(t \leq t_{critical}\), we fail to reject \(H_{0}\).

Conclusion

Since \(1.501 < 1.895\), we fail to reject the null hypothesis \(H_{0}\) at the \(\alpha = 0.05\) significance level.

Thus, there is not enough evidence to support the alternative hypothesis \(H_{1}: \mu_d > 0\).