Before the furniture store began its ad campaign, it averaged 159 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: 191, 161, 179, 160, 166, 181, 168, 163, 180, 164, 191, 147Assuming 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 z-test for a population proportion t-test for a population mean The null and alternative hypotheses would be: H0:H0: ? p μ Select an answer = > < ≠ H1:H1: ? p μ 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 ? ≤ > ααBased on this, we should Select an answer fail to reject accept 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 159 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 159.The data suggest the population mean is not significantly different from 159 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 159.The data suggest the populaton mean is significantly different from 159 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 159.Interpret the p-value in the context of the study.If the population mean number of customers since the ad campaign began is 159 and if you collect data for another 12 days since the ad campaign began, then there would be a 1.06605776% chance that the population mean would either be less than 147.1 or greater than 170.9.There is a 1.06605776% chance that the population mean number of customers since the ad campaign began is not equal to 159.If the population mean number of customers since the ad campaign began is 159 and if you collect data for another 12 days since the ad campaign began, then there would be a 1.06605776% chance that the sample mean for these 12 days would either be less than 147.1 or greater than 170.9.There is a 1.06605776% chance of a Type I error.Interpret the level of significance in the context of the study.If the population mean number of customers since the ad campaign began is 159 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 different from 159.There is a 5% chance that the population mean number of customers since the ad campaign began is different from 159.If the population mean number of customers since the ad campaign began is different from 159 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 159.There is a 5% chance that there will be no customers since everyone shops online nowadays.

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MATLAB: An Introduction with Applications

6th Edition

ISBN:9781119256830

Author:Amos Gilat

Publisher:Amos Gilat

Chapter1: Starting With Matlab

Section: Chapter Questions

Problem 1P

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Before the furniture store began its ad campaign, it averaged 159 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:

191, 161, 179, 160, 166, 181, 168, 163, 180, 164, 191, 147

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 z-test for a population proportion t-test for a population mean
The null and alternative hypotheses would be:

H0:H0: ? p μ Select an answer = > < ≠

H1:H1: ? p μ 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 fail to reject accept 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 159 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 159.
- The data suggest the population mean is not significantly different from 159 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 159.
- The data suggest the populaton mean is significantly different from 159 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 159.
Interpret the p-value in the context of the study.
- If the population mean number of customers since the ad campaign began is 159 and if you collect data for another 12 days since the ad campaign began, then there would be a 1.06605776% chance that the population mean would either be less than 147.1 or greater than 170.9.
- There is a 1.06605776% chance that the population mean number of customers since the ad campaign began is not equal to 159.
- If the population mean number of customers since the ad campaign began is 159 and if you collect data for another 12 days since the ad campaign began, then there would be a 1.06605776% chance that the sample mean for these 12 days would either be less than 147.1 or greater than 170.9.
- There is a 1.06605776% chance of a Type I error.
Interpret the level of significance in the context of the study.
- If the population mean number of customers since the ad campaign began is 159 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 different from 159.
- There is a 5% chance that the population mean number of customers since the ad campaign began is different from 159.
- If the population mean number of customers since the ad campaign began is different from 159 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 159.
- There is a 5% chance that there will be no customers since everyone shops online nowadays.

Definition Definition Measure of central tendency that is the average of a given data set. The mean value is evaluated as the quotient of the sum of all observations by the sample size. The mean, in contrast to a median, is affected by extreme values. Very large or very small values can distract the mean from the center of the data. Arithmetic mean: The most common type of mean is the arithmetic mean. It is evaluated using the formula: μ = 1 N ∑ i = 1 N x i Other types of means are the geometric mean, logarithmic mean, and harmonic mean. Geometric mean: The nth root of the product of n observations from a data set is defined as the geometric mean of the set: G = x 1 x 2 ... x n n Logarithmic mean: The difference of the natural logarithms of the two numbers, divided by the difference between the numbers is the logarithmic mean of the two numbers. The logarithmic mean is used particularly in heat transfer and mass transfer. ln x 2 − ln x 1 x 2 − x 1 Harmonic mean: The inverse of the arithmetic mean of the inverses of all the numbers in a data set is the harmonic mean of the data. 1 1 x 1 + 1 x 2 + ...

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