Question

A random sample of 20 married women showed that the mean time spent on housework by them was 29.8 hours a week with a standard deviation of 6.7 hours. If the margin of error is 3.14, find a 95% confidence interval for the mean time spent on housework per week by all married women.

Answer 1

95% confidence interval for the mean time spent on housework per week by all married women.

( 26.66 , 32.94)

Explanation:

Step(i):-

Given random sample size 'n' = 20

Mean of the sample (x⁻ ) = 29.8 hours

Standard deviation of the sample (S) = 6.7

Given Margin of error = 3.14

Step(ii):-

95% confidence interval for the mean is determined by

`(x^(-) - t_(0.05) (S)/(√(n) ) , x^(-) +t_(0.05) (S)/(√(n) ))`

We know that margin of error is determined by

`M.E = (t_(0.05)XS.D )/(√(n) ) = 3.14`

Now 95% confidence interval for the mean time spent on housework per week by all married women.

`(29.8 - 3.14 , 29.8+3.14)`

( 26.66 , 32.94)