I AM RESUBMITTING THIS QUESTION FOR DIFFERENT PARTS. I AM AT THE SECOND NUMBER 2 QUESTION "P-VALUE =". Before the furniture store began its ad campaign, it averaged 160 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: 131, 127, 148, 143, 141, 153, 158, 145, 138, 163, 164, 148, 166, 154, 170, 133 Assuming that the distribution is normal, what can be concluded at the αα = 0.05 level of significance? For this study, we should use Select an answer t-test for a population mean z-test for a population proportion The null and alternative hypotheses would be: H0:H0: ? μ p Select an answer < = ≠ > H1:H1: ? p μ Select an answer = > ≠ < The test statistic ? t z = (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 accept fail to reject reject the null hypothesis. 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 160 at αα = 0.05, so there is insufficient evidence to conclude that the population mean number of customers since the ad campaign began is different from 160. The data suggest the population mean is not significantly different from 160 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 160. The data suggest the populaton mean is significantly different from 160 at αα = 0.05, so there is sufficient evidence to conclude that the population mean number of customers since the ad campaign began is different from 160. Interpret the p-value in the context of the study. If the population mean number of customers since the ad campaign began is 160 and if you collect data for another 16 days since the ad campaign began, then there would be a 0.39912906% chance that the population mean would either be less than 148.9 or greater than 171.1. There is a 0.39912906% chance that the population mean number of customers since the ad campaign began is not equal to 160. There is a 0.39912906% chance of a Type I error. If the population mean number of customers since the ad campaign began is 160 and if you collect data for another 16 days since the ad campaign began, then there would be a 0.39912906% chance that the sample mean for these 16 days would either be less than 148.9 or greater than 171.1. Interpret the level of significance in the context of the study. If the population mean number of customers since the ad campaign began is 160 and if you collect data for another 16 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 different from 160. If the population mean number of customers since the ad campaign began is different from 160 and if you collect data for another 16 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 160. 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 different from 160.
I AM RESUBMITTING THIS QUESTION FOR DIFFERENT PARTS. I AM AT THE SECOND NUMBER 2 QUESTION "P-VALUE =".
Before the furniture store began its ad campaign, it averaged 160 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:
131, 127, 148, 143, 141, 153, 158, 145, 138, 163, 164, 148, 166, 154, 170, 133
Assuming that the distribution is normal, what can be concluded at the αα = 0.05 level of significance?
- For this study, we should use Select an answer t-test for a population mean z-test for a population proportion
- The null and alternative hypotheses would be:
H0:H0: ? μ p Select an answer < = ≠ >
H1:H1: ? p μ Select an answer = > ≠ <
- The test statistic ? t z = (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 accept fail to reject reject the null hypothesis.
- 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 160 at αα = 0.05, so there is insufficient evidence to conclude that the population mean number of customers since the ad campaign began is different from 160.
- The data suggest the population mean is not significantly different from 160 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 160.
- The data suggest the populaton mean is significantly different from 160 at αα = 0.05, so there is sufficient evidence to conclude that the population mean number of customers since the ad campaign began is different from 160.
- Interpret the p-value in the context of the study.
- If the population mean number of customers since the ad campaign began is 160 and if you collect data for another 16 days since the ad campaign began, then there would be a 0.39912906% chance that the population mean would either be less than 148.9 or greater than 171.1.
- There is a 0.39912906% chance that the population mean number of customers since the ad campaign began is not equal to 160.
- There is a 0.39912906% chance of a Type I error.
- If the population mean number of customers since the ad campaign began is 160 and if you collect data for another 16 days since the ad campaign began, then there would be a 0.39912906% chance that the sample mean for these 16 days would either be less than 148.9 or greater than 171.1.
- Interpret the level of significance in the context of the study.
- If the population mean number of customers since the ad campaign began is 160 and if you collect data for another 16 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 different from 160.
- If the population mean number of customers since the ad campaign began is different from 160 and if you collect data for another 16 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 160.
- 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 different from 160.
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Since there are multiple sub-parts, according to our policy we will solve the first three sub-parts.
You are at #2 ( P-value), so I am solving question numbers: 2, 3, and 4 for you.
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The manager is investigating if the average has changed since the ad came out and the data for 16 random days is given.
Also, the population standard deviation is unknown. Hence T-test would be more appropriate.
The hypothesis would be a two-tailed and test statistic value would be .-3.395891.
The P-value is required here. It can be easily found using MS-Excel.
Here, the test statistic ( t ) = -3.395891 and degrees of freedom = n- 1 = 16-1 = 15.
Also, it is the two-tailed test.
The Syntax/command for the p-value would be T.DIST.2T(test statistic, degrees of freedom)
Therefore to get a p-value, please type =T.DIST.2T(-3.395891,15) in an empty cell and press ENTER.
The p-value would be 0.003991 and till 4 decimal point, it would be 0.0040.
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