Data were collected on annual personal time (in hours) taken by a random sample of 16 women and 7 men employed by a medium sized company. The women took an average of 24.75 hours of personal time per year with a standard deviation of 2.84 hours. The men took an average of 21.89 hours of personal time per year with a standard deviation of 3.29 hours. The correct 90% confidence interval for the difference between women and men in the average number of hours of personal time taken per year is .

The correct 90% confidence interval for the difference between women and men in the average number of hours of personal time taken per year is

contained by the interval (0.54 , 5.18)

Step-by-step explanation:

Given that:

for women

sample size  = n₁ = 16

sample mean = 24.75

standard deviation s₁ = 2.84

for men

sample size for men = n₂ = 7

sample mean = 21.89

standard deviation s₂ = 3.29

degree of freedom =

= 16 + 7 -2

= 21

The critical value for

The pooled standard deviation can be calculated as follows since the population variances are assumed to be equivalent.

The standard error is estimated as follow:

se = 1.348

Finally, the confidence interval

sinn is the total amount of students now and m were the students before, you subtract them both (n-m) and divide that by m. then, since to get a percentage from a fraction you have to multiply by 100, your answer should look like this: n-m/m*100

step-by-step explanation: