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

Answer:

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

( 26.66 , 32.94)

Step-by-step 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

$(x^{-} - t_{0.05} \frac{S}{\sqrt{n} } , x^{-} +t_{0.05} \frac{S}{\sqrt{n} })$

We know  that margin of error is determined by

$M.E = \frac{t_{0.05}XS.D }{\sqrt{n} } = 3.14$

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)