Question

The difference of the sample means of two populations is 34.6, and the standard deviation of the difference of the sample means is 11.9. The 95% confidence interval lies between ______ and __________:

First BLANK:
A) -11.9
B) -23.8
C) -35.7
D) -45.4
Second BLANK:
A) +11.9
B) +23.8
C) +35.7
D) +45.4

Answer 1

The 95% confidence interval for the difference between the sample means, given a mean difference of 34.6 and standard deviation of 11.9, is approximately (11.276, 57.924). We calculate this using the formula involving the z-score for 95% confidence which is roughly 1.96.

Step-by-step explanation:

The question deals with finding the 95% confidence interval for the difference between two sample means, given the sample mean difference and the standard deviation of this difference. To find this confidence interval, typically a z-score is used for normally distributed data when the sample size is large, or the population standard deviation is known. For a 95% confidence interval and a normal distribution, we use a z-score that corresponds to the middle 95%, which is approximately 1.96 (this comes from the fact that 95% of data in a normal distribution lies within roughly 1.96 standard deviations from the mean in both directions).

Using the provided numbers, we calculate the confidence interval as follows:

Mean difference = 34.6

Standard deviation of the mean difference (SD) = 11.9

Margin of error (ME) = z-score * SD = 1.96 * 11.9

Lower limit of CI = Mean difference - ME
Upper limit of CI = Mean difference + ME

After plugging in the values and calculating:

1. Margin of error (ME) = 1.96 * 11.9 = 23.324
2. Lower limit = 34.6 - 23.324 = 11.276
3. Upper limit = 34.6 + 23.324 = 57.924

Therefore, the 95% confidence interval for the difference between the two sample means is approximately (11.276, 57.924).