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.

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

We know  that margin of error is determined by

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

( 26.66 , 32.94)

because 1.2 isnt half of 6 duh

so firstly, we need to isolate the y variables to be able to solve these inequalities. firstly, add 0.5x on both sides of the first inequality and subtract 1.5x on both sides of the second inequality:

now since the slope is positive for the first inequality, this means that the line going upwards belongs to the first inequality, and the line going downwards belongs to the second inequality.

next, since y ≥ in the first inequality, this means that the shaded region will be above the first inequality's line, thus shading regions a and b.

next, since y ≤ in the second inequality, this means that the shaded region is going to be below this line, thus shading regions a and c.

since both lines shade region a, this means that region a is the solution.

step-by-step explanation:

mark brainliest!