Question

Sarah Worker is making a presentation. She has 81 people in a workshop. She selects 64. The average budget expense for groceries is 8%. Sarah wants her workshop group to calculate a 95% confidence interval for this estimated proportion. The confidence interval is from 2% to 14% (to the nearest percent).
TRUE OR FALSE???

Answer 1

True is the answer hope i help

Answer 2

Sample size selected for survey = S= $\frac{64}{81}$=0.790

Z value for 95 % confidence interval = 1.96

Formula for calculating 95% confidence interval for this estimated proportion that is when 64 people are selected among 81 people:

     =  $S \pm {(\text{Z value)}\sqrt{\frac{S \times (1-S)}{N}}$

Where, S= Sample Size selected for survey=It's Probability=0.790

N= Total Size of Sample= 81

95% confidence interval for this estimated proportion

         = $0.790 \pm 1.96 \sqrt{\frac{0.79 \times (1-0.79)}{81}}\\\\ =0.790 \pm 1.96 \sqrt{\frac{0.79 \times (0.21)}{81}}\\\\ 0.790 \pm 0.08870$

= 0.790+ 0.08870 and 0.79 - 0.088

= 0.87 and 0.70(approx)

=Between( 70 % and 87 %)

→95 % Confidence interval for this estimated proportion that is selecting 64 out of  81 lies between= 70 % and 87 %

So,the statement that 95 % Confidence interval for this estimated proportion that is selecting 64 out of  81 lies between  2% to 14% is False.