The foot lengths of a random sample of 63 men had a mean of 28 cm and a standard deviation of 2.8 cm, while a random sample of 90 women had a mean foot length of 24.5 cm and a standard deviation of 2.8 cm. Compute an approximate 95% confidence interval for the difference between the mean foot lengths of men and women. (Round all answers to the nearest tenth.)
__?__ to __?__cm
To compute an approximate 95% confidence interval for the difference between the mean foot lengths of men and women, we can use the formula:
CI = (X1 - X2) ± z * sqrt((s1^2 / n1) + (s2^2 / n2))
Where:
- X1 and X2 are the means of the foot lengths of men and women, respectively.
- s1 and s2 are the standard deviations of the foot lengths of men and women, respectively.
- n1 and n2 are the sample sizes of men and women, respectively.
- z is the z-score corresponding to the desired confidence level.
The given information is:
- X1 = 28 cm (mean foot length of men)
- X2 = 24.5 cm (mean foot length of women)
- s1 = s2 = 2.8 cm (standard deviations of both groups)
- n1 = 63 (sample size of men)
- n2 = 90 (sample size of women)
- The desired confidence level is 95%, so the corresponding z-score is 1.96.
Now, let's substitute the values into the formula:
CI = (28 - 24.5) ± 1.96 * sqrt((2.8^2 / 63) + (2.8^2 / 90))
Calculating the expression inside the square root:
CI = (28 - 24.5) ± 1.96 * sqrt((0.1111) + (0.07134))
CI = (28 - 24.5) ± 1.96 * sqrt(0.18244)
CI = (28 - 24.5) ± 1.96 * 0.4273
CI = 3.5 ± 0.8374
Now, we can calculate the upper and lower bounds of the confidence interval:
Upper bound = 3.5 + 0.8374
Lower bound = 3.5 - 0.8374
Upper bound ≈ 4.3374
Lower bound ≈ 2.6626
Therefore, the approximate 95% confidence interval for the difference between the mean foot lengths of men and women is __2.7 cm__ to __4.3 cm__. (Round to the nearest tenth)