Question

Before the furniture store began its ad campaign, it averaged 193 customers per day. The manager is investigating if the average is smaller since the ad came out. The data for the 12 randomly selected days since the ad campaign began is shown below:

200, 181, 185, 205, 198, 201, 173, 165, 180, 198, 199, 200

Assuming that the distribution is normal, what can be concluded at the αα = 0.05 level of significance?

1. For this study, we should use Select an answer z-test for a population proportion t-test for a population mean
2. The null and alternative hypotheses would be:

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

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

3. The test statistic ? z t  =  (please show your answer to 3 decimal places.)
4. The p-value =  (Please show your answer to 4 decimal places.)
5. The p-value is ? > ≤  αα
6. Based on this, we should Select an answer reject accept fail to reject  the null hypothesis.
7. Thus, the final conclusion is that ...
   • The data suggest the populaton mean is significantly less than 193 at αα = 0.05, so there is sufficient evidence to conclude that the population mean number of customers since the ad campaign began is less than 193.
   • The data suggest the population mean is not significantly less than 193 at αα = 0.05, so there is sufficient evidence to conclude that the population mean number of customers since the ad campaign began is equal to 193.
   • The data suggest that the population mean number of customers since the ad campaign began is not significantly less than 193 at αα = 0.05, so there is insufficient evidence to conclude that the population mean number of customers since the ad campaign began is less than 193.
8. Interpret the p-value in the context of the study.
   • There is a 25.3512784% chance that the population mean number of customers since the ad campaign began is less than 193.
   • There is a 25.3512784% chance of a Type I error.
   • If the population mean number of customers since the ad campaign began is 193 and if you collect data for another 12 days since the ad campaign began, then there would be a 25.3512784% chance that the sample mean for these 12 days would be less than 190.4.
   • If the population mean number of customers since the ad campaign began is 193 and if you collect data for another 12 days since the ad campaign began, then there would be a 25.3512784% chance that the population mean number of customers since the ad campaign began would be less than 193.
9. Interpret the level of significance in the context of the study.
   • If the population mean number of customers since the ad campaign began is 193 and if you collect data for another 12 days since the ad campaign began, then there would be a 5% chance that we would end up falsely concuding that the population mean number of customers since the ad campaign began is less than 193.
   • If the population mean number of customers since the ad campaign began is less than 193 and if you collect data for another 12 days since the ad campaign began, then there would be a 5% chance that we would end up falsely concuding that the population mean number of customers since the ad campaign is equal to 193.
   • There is a 5% chance that there will be no customers since everyone shops online nowadays.
   • There is a 5% chance that the population mean number of customers since the ad campaign began is less than 193.

Answer 1

a) In this study we will use the t test for population mean.

b) Null hypothesis: There is no difference in average number of customers arriving the furniture store before and after the ad campaign.

Alternative hypothesis: The ad campaign increased the average number of customers arriving at the furniture store

c) t statistic is given by

Where xbar is sample mean = 190.42 and mu is specified value of mean before ad campaign which is 193

Since standard deviation of population is unknown we cam replace it by sample standard deviation using formula √[ summation( xi - xbar)square / n-1] = 13.05

So, t = 2.58/3.73 = 0.691

Now critical value of the t distribution at 5% level of significance and 11 degree of freedom is 2.201.

Since calculated t is less than critical value of t, null hypothesis may not be rejected. So there is no difference in number of customers before and after the ad campaign.

d) p value is 0.503887 which is greater than 0.05, hence in favor of the null hypothesis.