Question

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 if 99% confidence is desired?

Answer 1

3.4779 .

Explanation:

Answer 2

Answer:
`3.4779`

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 .