Data were collected on the amount spent for lunch by 64 customers at a major Houston...

Data were collected on the amount spent for lunch by 64 customers at a major Houston restaurant. The sample provided a sample mean of $21.5. Based upon past studies the population standard deviation is known with σ = $6. Develop a 99% confidence interval estimate of the mean amount spent for lunch.

Solutions

Expert Solution

Solution :

Given that,

Sample size = n = 64

Z/2 = 2.576

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

= 2.576 * (6 / $\sqrt$ 64)

= 1.93

At 99% confidence interval estimate of the population mean is,

- E < < + E

21.5 - 1.93 < < 21.5 + 1.93

19.57 < < 23.43

( 19.57, 23.43)

orchestra answered 1 year ago

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