Question

# Hello, something I'm doing on Excel is wrong and need to know how to do this...

Hello, something I'm doing on Excel is wrong and need to know how to do this correctly on Excel, now written out.

A sample of the sales at 50 Haute Dog restaurants is taken and the sample mean is \$30,000 per day

with a sample standard deviation of \$2000. The population standard deviation is not known.

a. What is the 90% confidence interval for the daily sales at all restaurants.

b. What is the probability that the average sales are really \$32,000 or more per day?

Assume that the population average is \$30,000 per day and the std. Deviation is \$2k.

Solution:

We are given n=50, Xbar = 30000, S = 2000

Part a

Confidence interval for population mean is given as below:

Confidence interval = Xbar ± t*S/sqrt(n)

Confidence level = 90%

df = n – 1 = 50 – 1 = 49

Critical t value = 1.6766

(by using excel command =TINV(1 – 0.90, 49))

Confidence interval = 30000 ± 1.6766*2000/sqrt(50)

Confidence interval = 30000 ± 1.6766* 282.8427125

Confidence interval = 30000 ± 474.2002

Lower limit = 30000 - 474.2002 = 29525.80

Upper limit = 30000 + 474.2002 = 30474.20

Confidence interval = (\$29525.80, \$30474.20)

Part b

Here, we have to find P(Xbar>32000)

Z = (Xbar - µ) / [σ/sqrt(n)]

Z = (32000 – 30000) / [2000/sqrt(50)]

Z = 2000/[2000/7.071068]

Z = 2000/282.8427

Z = 7.071068

P(Z<7.071068) = 1.00 (by using z-table/excel)

(Use excel command =normsdist(7.071068))

P(Z>7.071068) = 0.00

P(Xbar>32000) = 0.00

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