Question

The following sample of eight matched pairs
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.

Answers

Answer
(b) Compute the test statistic. Round your final answer to at least three decimal places.
Answer
Can you reject Hv0 at the a-0.05 level
Answered by GPT-4o mini
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: \(2.375\)
Answered by GPT-4o mini
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.
Answered by GPT-4o mini
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 \(H_{1}: \mu_d > 0\).

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