Question

Before the furniture store began its ad campaign, it averaged 103 customers per day. The manager is investigating if the average has changed since the ad came out. The data for the 16 randomly selected days since the ad campaign began is shown below:

108, 89, 104, 75, 111, 84, 88, 71, 96, 93, 107, 107, 69, 109, 97, 90

Assuming that the distribution is normal, what can be concluded at the αα = 0.01 level of significance?

For this study, we should use Select an answer z-test for a population proportion t-test for a population mean

The null and alternative hypotheses would be:

H0:H0:  ? μ p  Select an answer < > = ≠       

H1:H1:  ? μ p  Select an answer = > ≠ <    

The test statistic ? z t  =  (please show your answer to 3 decimal places.)

The p-value =  (Please show your answer to 4 decimal places.)

The p-value is ? ≤ >  αα

Based on this, we should Select an answer reject accept fail to reject  the null hypothesis.

Thus, the final conclusion is that ...

The data suggest the populaton mean is significantly different from 103 at αα = 0.01, so there is sufficient evidence to conclude that the population mean number of customers since the ad campaign began is different from 103.

The data suggest that the population mean number of customers since the ad campaign began is not significantly different from 103 at αα = 0.01, so there is insufficient evidence to conclude that the population mean number of customers since the ad campaign began is different from 103.

The data suggest the population mean is not significantly different from 103 at αα = 0.01, so there is sufficient evidence to conclude that the population mean number of customers since the ad campaign began is equal to 103.

Interpret the p-value in the context of the study.

If the population mean number of customers since the ad campaign began is 103 and if you collect data for another 16 days since the ad campaign began, then there would be a 1.5820853% chance that the sample mean for these 16 days would either be less than 93.6 or greater than 112.4.

If the population mean number of customers since the ad campaign began is 103 and if you collect data for another 16 days since the ad campaign began, then there would be a 1.5820853% chance that the population mean would either be less than 93.6 or greater than 112.4.

There is a 1.5820853% chance of a Type I error.

There is a 1.5820853% chance that the population mean number of customers since the ad campaign began is not equal to 103.

Interpret the level of significance in the context of the study.

There is a 1% chance that the population mean number of customers since the ad campaign began is different from 103.

There is a 1% chance that there will be no customers since everyone shops online nowadays.

If the population mean number of customers since the ad campaign began is 103 and if you collect data for another 16 days since the ad campaign began, then there would be a 1% chance that we would end up falsely concuding that the population mean number of customers since the ad campaign began is different from 103.

If the population mean number of customers since the ad campaign began is different from 103 and if you collect data for another 16 days since the ad campaign began, then there would be a 1% chance that we would end up falsely concuding that the population mean number of customers since the ad campaign is equal to 103.

Answer 1

null hypothesis: HO: μ	=	103
Alternate Hypothesis: Ha: μ	≠	103

sample mean 'x̄=	93.625
sample size   n=	16.00
sample std deviation s=	13.79
std error 'sx=s/√n=	3.4471

test stat t ='(x-μ)*√n/sx=

-2.720

p value      =

0.0158

The p-value is > alpha:

The data suggest that the population mean number of customers since the ad campaign began is not significantly different from 103 at αα = 0.01, so there is insufficient evidence to conclude that the population mean number of customers since the ad campaign began is different from 103

If the population mean number of customers since the ad campaign began is 103 and if you collect data for another 16 days since the ad campaign began, then there would be a 1.5820853% chance that the sample mean for these 16 days would either be less than 93.6 or greater than 112.4.