Question

4. The owner of specialty coffee shop wants to study coffee purchasing habits of customers at her shop. She selects a random sample of 60 customers during certain week, with the following results:

The amount spent was ? = $7.25, S = $1.75

Thirty-one customers say they definitely will” recommend the specialty coffee shop to family friends.

a. At the 0.05 level of significance, is there evidence that the population mean amount spent was different from $6.50? (IS THIS RIGHT?)

H0:=6.50

H: doesnt equal 6.50

Population standard deviation is not given, sample size is large, using sample t confidence interval for population mean.

n=60

x=7.25

s=1.75

SE Mean=0.226

95% confidence interval=(6.798,7.702)

T=3.32

b. Determine the p-value in (a). (IS THIS RIGHT?)

P-Value=.002

Since P-Value is is less than .05 the null hypothesis is rejected.

c. At the 0.05 level of significance, is there evidence that more than 50% of all customers say they “definitely will” recommend the specialty coffee shop to family and friends? (IS THIS RIGHT?)

With a 5% significance level there is evidence that the population mean amount spent was different from the $6.50.

d. What is your answer to (a) if the sample mean equals $6.25? (HELP)

e. What is your answer to (c) if 39 customers say they “definitely will” recommend the specialty coffee to family and friends? (HELP)

f. Submit an excel file which clearly describes the data, intermediate calculations, and final results. Also submit a word file to provide explanation of each step while solving the problems. (HELP)

Answer 1

a) Null Hypothesis

Alternative Hypothesis

Under H0, the test statistic is

Degrees of freedom = n-1= 60-1 = 59

b)  The P-Value is .001547

Since P-Value is is less than .05 the null hypothesis is rejected.

c) Null Hypothesis

Alternative Hypothesis

The sample proportion of customers say they “definitely will” recommend the specialty coffee shop to family and friends is

Under H0, the test statistic is

The P-Value is 0.3979

Since p value is greater than significance level, Fail to Reject H0.

Hence, there is not sufficient evidence to conclude that more than 50% of customers say they "definitely will" recommend the specialty coffee shop to family and friends.

d)

Null Hypothesis

Alternative Hypothesis

Under H0, the test statistic is

Degrees of freedom = n-1= 60-1 = 59

Significance level = 0.05

The P value is 0.2730

Since the p-value is greater than significance level , Fail to reject H0.

Hence, there is not enough evidence to conclude that the mean amount spent differs from $6.50

e)

Null Hypothesis

Alternative Hypothesis

The sample proportion of customers say they “definitely will” recommend the specialty coffee shop to family and friends is

Under H0, the test statistic is

The P-Value is 0.0101

Since p value is less than significance level, Reject H0.

Hence, there is sufficient evidence to conclude that more than 50% of customers say they "definitely will" recommend the specialty coffee shop to family and friends.