A survey of 10 resturants in a fast food resturant group showed a mean customer satisfaction index of 73 with a standard deviation of the index is 6. What is the margin of error…

Question

asked Dec 12, 2020 179k views

2 Answers

Gerichhome · Answer 1 · 2020-12-17T21:10:08+0000

Answer:

Explanation:

Given : Sample size :
n=10

Sample mean :
$\overline{x}=73$

Standard deviation :
$\sigma =6$

Significance level :
$\alpha=1-0.99=0.01$

Critical value :
$t_(n-1,\alpha/2)=t_(9,0.005)=1.833$

Formula to find the margin of error for population mean :-

$t_(n-1,\alpha/2)(\sigma)/(√(n))\\\\=(1.833)(6)/(√(10))\\\\\approx3.4779$

Hence, the margin of error if 99% confidence is 3.4779 .