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A restaurant wants to estimate the average bill paid by a party of two for dinner...

A restaurant wants to estimate the average bill paid by a party of two for dinner (service hours after 5 p.m.) They take a random sample of 100 bills and find that they average $80 with a standard deviation of $40. In making an 80% confidence interval for the parameter of interest, they should get a margin of error of about(Answer)?

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Expert Solution

Solution :

Given that,

Point estimate = sample mean = = 80

Population standard deviation = = 40

Sample size = n = 100

At 80% confidence level the z is ,

= 1 - 80% = 1 - 0.80 = 0.20

/ 2 = 0.20 / 2 = 0.10

Z/2 = Z0.10 = 1.28

Margin of error = E = Z/2* ( /n)

= 1.28 * (40 / 100)

= 5.12

Margin of error = E = 5.12

At 80% confidence interval estimate of the population mean is,

- E < < + E

80 - 5.12 < < 80 + 5.12

74.88 < < 85.12

(74.88, 85.12 )

milcah answered 4 days ago
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