Question

ASAP HELP!!

You are trying to determine if it matters what check out line you are in at Cost-Co. You randomly observe 368 shoppers as they check-out at Cost-Co. Four check-out lines are open. The results are displayed in the table below. Use a level of significance of a =0.01. a. Complete the rest of the table by filling in the expected frequencies: Frequency of Shoppers in Each Line Check-Out Line Frequency Expected Frequency Line 1 82 Line 2 102 Line 3 69 Line 4 115 b. What is the correct statistical test to use? Select an answer c. What are the null and alternative hypotheses? H: The shoppers and lines are independent. The distribution of shoppers is uniform. The shoppers and lines are dependent The distribution of shoppers is not uniform. H: The shoppers and lines are dependent. The distribution of shoppers is uniform. The shoppers and lines are independent. The distribution of shoppers is not uniform. d. The degrees of freedom =

Answer 1

The given data is about the number of shoppers present in different lines. Specifically, there are four lines of shoppers. We are to determine if it matters as to which line one is present in. I am presenting the answer as a traditional hypothesis question, you can choose the appropriate answers based on these explicit explanations.

The alternate hypothesis would be that it would matter as to what line one is present in. That is to say that shoppers in each of the lines is not uniform but there are a different number of shoppers in each line.

The null hypothesis would be negation of this and hence would be that the number of shoppers in each of the lines is uniform.

In order to validate the null, we can perform a chi-square test. The significance level is given to be 0.01. There are a total of n = 4 lines. Therefore, the degrees of freedom are df = n - 1 = 4 - 1 = 3.

The observed data is:

Check-Out Line	Observed Frequency
Line 1	82
Line 2	102
Line 3	69
Line 4	115
TOTAL	368

Next we are to determine the expected number of shoppers in each line. This is to be done assuming the null which states that the number of shoppers in each line is uniform. Therefore, the number of shoppers in each line would be 368 / 4 = 92.

Check-Out Line	Observed Frequency	Expected Frequency
Line 1	82	92
Line 2	102	92
Line 3	69	92
Line 4	115	92
TOTAL	368	368

Next we calculate the ratio:

$\frac{(Observed - Expected)^2}{Expected}$

Check-Out Line	Observed Frequency	Expected Frequency	$\frac{(Observed - Expected)^2}{Expected}$
Line 1	82	92	1.087
Line 2	102	92	1.087
Line 3	69	92	5.75
Line 4	115	92	5.75
TOTAL	368	368	13.674

The chi-square statistic is given using the formula:

$\chi^2 = \sum \frac{(Observed - Expected)^2}{Expected}$

Summing the last column, we get:

$\chi^2 = 13.674$

Using the aforementioned significance level of 0.01, and df = 3, we get a p-value of p = 0.0034.

Clearly, p = 0.0034 < 0.01. Therefore, the p-value is smaller than the significance level. Therefore, we can reject the null hypothesis and accept the alternate hypothesis.

Therefore we conclude that there is sufficient evidence that the distribution of shoppers is not uniform.

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