Question

A New York restaurant would like to estimate the average amount customers spend for dinner. A sample of size 49 customers resulted an average of $32.60. Assume the population standard deviation is $5.What is the point estimation for the mean of the population?At 90% level of confidence, what is the margin of error?What is the 90% confidence level for the mean of the population?

Answer 1

Answer: Point of estimation = 32.60

Margin of error = 1.175.

95% confidence interval would be (81.95,84.45).

Explanation:

Since we have given that

n = 49

Average = $32.60

Standard deviation = $5

We need to find:

1) Point of estimation for the mean of the population:

`\mu=\$32.60`

2) At 90% level of confidence, z = 1.645

Margin of error would be

`z(\sigma)/(√(n))\\\\=1.645* (5)/(√(49))\\\\=1.645* (5)/(7)\\\\=1.175`

So, Margin of error = 1.175.

3) At 90% level of confidence, interval would be

`\bar{x}\pm 1.175\\\\=32.60\pm 1.175\\\\=(32.60-1.175,32.60+1.175)\\\\=(31.425,33.775)\\\\=(31.43,33.76)`

Hence, 90% confidence interval would be (31.43,33.76).