Question

Before the furniture store began its ad campaign, it averaged 147 customers per day. The manager...

Before the furniture store began its ad campaign, it averaged 147 customers per day. The manager is investigating if the average has changed since the ad came out. The data for the 12 randomly selected days since the ad campaign began is shown below:

147, 128, 147, 149, 127, 137, 157, 139, 146, 151, 133, 124

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

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

b. The null and alternative hypotheses would be:   

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

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

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

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

e. The p-value is? ≤ > α

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

g. Thus, the final conclusion is that ...

The data suggest that the population mean number of customers since the ad campaign began is not significantly different from 147 at α = 0.01, so there is insufficient evidence to conclude that the population mean number of customers since the ad campaign began is different from 147.

The data suggest the population mean is significantly different from 147 at α = 0.01, so there is sufficient evidence to conclude that the population mean number of customers since the ad campaign began is different from 147.

The data suggest the population mean is not significantly different from 147 at α = 0.01, so there is sufficient evidence to conclude that the population mean number of customers since the ad campaign began is equal to 147.

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

If the population mean number of customers since the ad campaign began is 147 and if you collect data for another 12 days since the ad campaign began, then there would be a 5.57459492% chance that the population mean would either be less than 140.4 or greater than 153.6.

There is a 5.57459492% chance that the population mean number of customers since the ad campaign began is not equal to 147.

If the population mean number of customers since the ad campaign began is 147 and if you collect data for another 12 days since the ad campaign began, then there would be a 5.57459492% chance that the sample mean for these 12 days would either be less than 140.4 or greater than 153.6.

There is a 5.57459492% chance of a Type I error.

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

If the population mean number of customers since the ad campaign began is different from 147 and if you collect data for another 12 days since the ad campaign began, then there would be a 1% chance that we would end up falsely concluding that the population mean number of customers since the ad campaign is equal to 147.

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

There is a 1% chance that the population mean number of customers since the ad campaign began is different from 147.

If the population mean number of customers since the ad campaign began is 147 and if you collect data for another 12 days since the ad campaign began, then there would be a 1% chance that we would end up falsely concluding that the population mean number of customers since the ad campaign began is different from 147.

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Answer #1

a. For this study, we should use t-test for a population mean

because the manager wants to investigate for the average.

b. The null and alternative hypotheses would be:   

H0: μ = 147   

because this is the initial situation.

H1: μ ≠ 147

because the manager wants to investigate if the average has changed. and change ca be both greater than or less than to include both possibilities we use not equal to.

Hence, two tailed test will be reuqired.

c. The test statistic? t  = ( -2.164)  (please show your answer to 3 decimal places.)

We will use ti 84 for this.

1. Press [STAT] then go the TESTS menu.

EDIT CALC TESTS 18Z-Test... ZA T-Test... 3:2-SAMPZTest.... 4:2-SAMPTTest.... 5:1-PropZTest.... 6:2-PropZTest... 7+ZInterval..

2. Select “2. T-test”. Make sure that you highlight Data and press [ENTER] if your screen looks different from this.

T-Test Inpt:Data Stats WO: Sx: 0 n: piwo <HO SHO Calculate Draw

ensure you enter the data in the list L1

L3 L1 L2 142 128 147 149 127 137 157 L101)=147

3. Enter the values and select the correct tail for the test.

T-Test Inpt:Datz Stats 10:147 List: L1 Fre: 10 .: N.o <HO SHO Calculate Draw

T-Test #147 t=-2.164043325 P=.053327546 X=140.25 Sx=10.80509139 n=12

d. The p-value = 0.0533 (Please show your answer to 4 decimal places.)

e. The p-value is 0.05 > α (=0.01)

f. Based on this, we should fail to reject the null hypothesis. as p-value > alpha value.

As per the HOMEWORKLIB POLICY we can answer only the first 4 parts.

If it helps, kindly upvote.

Comment below for doubt.

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