Question

1. Collette is self-employed and sells cookware at home parties. She wants to estimate the average amount a client spends at each party. A random sample of 35 receipts gave a mean of = $34.70 with standard deviation s = $4.85.

1. a. Find a 90% confidence interval for the average amount spent by all clients. Round your final answers to the nearest hundredth.

b. Interpret the meaning of the confidence interval in the context of this problem. Include units in your answer.

1. c. Collette needs the customers to spend an average of $40 to continue her business. Do they spend enough? Why or why not?

Answer 1

Solution-A:

alpha=1-0.90=0.10

alpha/2=0.10/2=0.05

df=n-1=35-1=34

t critical in excel is

=T.INV(0.05,34)

=1.690924255

90% confidence interval for mean is

xbar-t*s/sqrt(n),xbar+t*s/sqrt(n)

34.70-1.690924255*4.85/sqrt(35),34.70+1.690924255*4.85/sqrt(35)

33.31378, 36.08622

33.31,36.09

. Interpret the meaning of the confidence interval in the context of this problem. Include units in your answer.

we are 90% confident that the true average amount spent by all clients lies in between $ 33.31 and $ 36.09

Solution:

Ho:mu=40

Ha:mu not = 40

since mu=40 is not in the 90% confidence interval range

Reject Ho

Accept Ha

since average is different from 40

They dont spend enough.