Question

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.

Answer 1

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