Given two dependent random samples with the following results:
| Population 1 | 30 | 47 | 19 | 49 | 42 | 31 | 24 |
|---|---|---|---|---|---|---|---|
| Population 2 | 45 | 37 | 29 | 44 | 46 | 46 | 34 |
Use this data to find the $90 \%$ confidence interval for the true difference between the population means. Assume that both populations are normally distributed.
Step 1 of 4: Find the point estimate for the population mean of the paired differences. Let $x_1$ be the value from Population 1 and $x_2$ be the value from Population 2. Use the formula $d=x_2-x_1$ to calculate the paired differences. Round your answer to one decimal place.
Answer (1)
Calculate the paired differences: d = x 2 − x 1 .
Find the mean of the differences: d ˉ ≈ 5.6 .
Determine the 90% confidence interval: ( − 1.61 , 12.75 ) .
The point estimate for the population mean of the paired differences is 5.6 .
Explanation
Calculate Paired Differences We are given two dependent random samples and asked to find the 90% confidence interval for the true difference between the population means. We are also asked to find the point estimate for the population mean of the paired differences. The first step is to calculate the paired differences d = x 2 − x 1 for each pair of data points.
Calculate the Differences Let's calculate the paired differences:
d 1 = 45 − 30 = 15 d 2 = 37 − 47 = − 10 d 3 = 29 − 19 = 10 d 4 = 44 − 49 = − 5 d 5 = 46 − 42 = 4 d 6 = 46 − 31 = 15 d 7 = 34 − 24 = 10
So, the paired differences are: {15, -10, 10, -5, 4, 15, 10}.
Calculate the Mean of Differences Next, we calculate the sample mean of the paired differences, denoted as d ˉ .
d ˉ = n ∑ d i = 7 15 + ( − 10 ) + 10 + ( − 5 ) + 4 + 15 + 10 = 7 39 ≈ 5.5714 Rounding to one decimal place, we get d ˉ ≈ 5.6 .
Calculate Standard Deviation of Differences Now, we calculate the sample standard deviation of the paired differences, denoted as s d .
First, we calculate the squared differences from the mean ( d i − d ˉ ) 2 :
( 15 − 5.57 ) 2 ≈ 88.95 ( − 10 − 5.57 ) 2 ≈ 242.36 ( 10 − 5.57 ) 2 ≈ 19.62 ( − 5 − 5.57 ) 2 ≈ 111.72 ( 4 − 5.57 ) 2 ≈ 2.46 ( 15 − 5.57 ) 2 ≈ 88.95 ( 10 − 5.57 ) 2 ≈ 19.62
Sum of squared differences: 88.95 + 242.36 + 19.62 + 111.72 + 2.46 + 88.95 + 19.62 = 573.68
Then, we calculate the sample variance: s d 2 = n − 1 ∑ ( d i − d ˉ ) 2 = 7 − 1 573.68 = 6 573.68 ≈ 95.61
Finally, we calculate the sample standard deviation: s d = s d 2 = 95.61 ≈ 9.7785
Find the t-critical Value We need to find the critical value t α /2 for a 90% confidence interval with n − 1 = 7 − 1 = 6 degrees of freedom. For a 90% confidence interval, α = 1 − 0.90 = 0.10 , so α /2 = 0.05 . We look up the t-value for 6 degrees of freedom and a 0.05 significance level in a t-table or use a calculator. The t-critical value is approximately 1.943.
Calculate the Margin of Error Now, we calculate the margin of error E using the formula: $E = t_{\alpha/2} * \frac{s_d}{\sqrt{n}} = 1.943 * \frac{9.7785}{\sqrt{7}} \approx 1.943 * \frac{9.7785}{2.6458} \approx 1.943 * 3.695 \approx 7.18
Calculate Confidence Interval Next, we calculate the lower and upper bounds of the confidence interval: Lower bound: d ˉ − E = 5.57 − 7.18 = − 1.61 Upper bound: d ˉ + E = 5.57 + 7.18 = 12.75
Therefore, the 90% confidence interval for the true difference between the population means is approximately (-1.61, 12.75).
State the Point Estimate The point estimate for the population mean of the paired differences is d ˉ ≈ 5.6 .
Examples
Consider a scenario where you want to test the effectiveness of a new drug designed to lower blood pressure. You measure the blood pressure of a group of patients before and after administering the drug. The paired difference is the difference in blood pressure for each patient (after - before). Calculating the confidence interval for the mean difference helps you determine if the drug has a statistically significant effect on blood pressure. If the confidence interval does not contain zero, it suggests that the drug has a significant impact. For example, if the 90% confidence interval is (-1.61, 12.75), it indicates that we are 90% confident that the true average change in blood pressure lies between -1.61 and 12.75 units.
