Question

As part of a competitive analysis, your group has just completed a small survey of (what you hope and assume are) randomly selected individuals from the population of potential customers you would like to learn about. You are particularly interested in their engagement with competitors, as this might represent a business opportunity for your firm. Results include reported time spent during the past week using competitors’ products as listed here in minutes:

              Time

                  70

                  94

                  10

                182

                100

                130

                    0

                160

                190

                  50

                270

                  93

                  10

a.   Find the average and the standard error of the average. Interpret this standard error.

Average:

Standard Error:

Interpretation of Standard Error:

b.   Find the 95% confidence interval for the population mean time.

Confidence Interval:

c.   The marketing group has concluded that this population would be worth pursuing more actively if engagement with competitors’ products in the population exceeds 30 minutes per week on average. Please report the results of the relevant two-sided test as indicated:

                              Significant?

Yes			p-Value:
No

d.   If your result in part c is significant, please state the one-sided conclusion to a significant two-sided test using numbers for the sample average and for the reference value. Otherwise (if not significant) please state the conclusion by indicating which hypothesis was accepted.

Answer 1

a)

mean of a data() is given by = 104.54

Sample standard deviation s =   = 80.44

For one mean, Standard error(SE) = s / sqrt(n) = 80.44/sqrt(13) = 22.31

The standard error tells us the deviation of the sample mean from the true mean. This is a large standard error, so it is expected that this sample mean 104.54 be very close to true mean or maybe very far like 22.31 units away.

b) 95% confidence interval =

For a significance level of 0.05, and degrees of freedom 13-1 = 12 is 2.18

So 95% confidence interval is (104.54 - 2.18*22.31, 104.54 + 2.18*22.31) = (55.90,153.17)

c)

Let's find t-statistic accordingly,

t = (X - 30)/SE = 104.54 - 30 / 22.31 = 3.34

For this t-statistic, and df = 12, p value comes out to be 0.005

Since p<0.05, hence we have strong evidence against the null hypothesis. It can also be confirmed by confidence interval too. Since 30 lies out of the confidence interval hence it the null hypothesis cannot be true for mean = 30.

Hence, it mean not equal to 30.

d) Result was significant, so here the null and alternate hypothesis would be

For this one-sided hypothesis, p-value comes out to be 0.003

Again null hypothesis is rejected and it can be claimed with 95% confidence that the mean is greater than 30.