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 and use the formula $d=x_2-x_1$ to calculate the paired differences. Round your answer to one decimal place.
Answer (2)
Calculate the paired differences: d = x 2 − x 1 .
Find the mean of the differences: d ˉ ≈ 5.6 .
Compute the standard deviation of the differences: s d ≈ 9.8 .
Determine the 90% confidence interval: ( − 1.6 , 12.8 ) .
Explanation
Problem Analysis We are given two dependent random samples and asked to find the 90% confidence interval for the true difference between the population means. We will calculate the paired differences, the mean and standard deviation of these differences, and then use the t-distribution to find the confidence interval.
Calculate Paired Differences First, we calculate the paired differences d = x 2 − x 1 :
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 mean of the paired differences, d ˉ :
d ˉ = 7 15 + ( − 10 ) + 10 + ( − 5 ) + 4 + 15 + 10 = 7 39 ≈ 5.6
Calculate the Standard Deviation of Differences Now, we calculate the sample standard deviation of the paired differences, s d :
First, we calculate the squared differences from the mean ( d i − d ˉ ) 2 :
( 15 − 5.6 ) 2 = 9. 4 2 = 88.36 ( − 10 − 5.6 ) 2 = ( − 15.6 ) 2 = 243.36 ( 10 − 5.6 ) 2 = 4. 4 2 = 19.36 ( − 5 − 5.6 ) 2 = ( − 10.6 ) 2 = 112.36 ( 4 − 5.6 ) 2 = ( − 1.6 ) 2 = 2.56 ( 15 − 5.6 ) 2 = 9. 4 2 = 88.36 ( 10 − 5.6 ) 2 = 4. 4 2 = 19.36
Sum of squared differences = 88.36 + 243.36 + 19.36 + 112.36 + 2.56 + 88.36 + 19.36 = 573.72
Then, we calculate the sample variance: s d 2 = n − 1 ∑ ( d i − d ˉ ) 2 = 7 − 1 573.72 = 6 573.72 = 95.62
Finally, the sample standard deviation is: s d = 95.62 ≈ 9.8
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. Since α = 1 − 0.90 = 0.10 , then α /2 = 0.05 . Using a t-table or calculator, we find that t 0.05 , 6 ≈ 1.943 .
Calculate the Margin of Error Now we calculate the margin of error E :
E = t α /2 ∗ n s d = 1.943 ∗ 7 9.8 ≈ 1.943 ∗ 2.646 9.8 ≈ 1.943 ∗ 3.704 ≈ 7.2
Calculate the Confidence Interval The lower bound of the confidence interval is: d ˉ − E = 5.6 − 7.2 = − 1.6
The upper bound of the confidence interval is: d ˉ + E = 5.6 + 7.2 = 12.8
Therefore, the 90% confidence interval for the true difference between the population means is ( − 1.6 , 12.8 ) .
Final Answer The 90% confidence interval for the true difference between the population means is approximately ( − 1.6 , 12.8 ) .
Examples
Understanding the confidence interval for the difference between two population means is useful in various fields. For example, a marketing team might want to know the difference in customer satisfaction scores before and after a promotional campaign. By calculating a confidence interval, they can be 90% confident that the true difference in satisfaction lies within a specific range, helping them make informed decisions about the campaign's effectiveness. Similarly, in healthcare, researchers might compare the effectiveness of two different treatments by examining the confidence interval of the difference in patient outcomes.
To find the point estimate for the population mean of the paired differences, we calculate the differences between pairs of values from two populations. The mean of these differences is found to be approximately 5.6, which serves as the point estimate. This value is rounded to one decimal place as requested.
;
