A market researcher working for the bank in Exercise 2 wants...

A market researcher working for the bank in Exercise 2 wants to know if the distribution of applications by card is the same for the past three mailings. She takes a random sample of 200 from each mailing and counts the number applying for Silver Gold and Platinum. The data follow: (FIGURE CAN'T COPY) a) What is the null hypothesis? b) What type of test is this? c) What are the expected numbers for each cell if the null hypothesis is true? d) Find the $\chi^2$ statistic. e) How many degrees of freedom does it have? f) Find the critical value at $\alpha=0.05$. g) What do you conclude?

A market researcher working for the bank in Exercise 2 wants to know if the distribution of applications by card is the same for the past three mailings. She takes a random sample of 200 from each mailing and counts the number applying for Silver Gold and Platinum. The data follow: (FIGURE CAN'T COPY)
a) What is the null hypothesis?
b) What type of test is this?
c) What are the expected numbers for each cell if the null hypothesis is true?
d) Find the $\chi^2$ statistic.
e) How many degrees of freedom does it have?
f) Find the critical value at $\alpha=0.05$.
g) What do you conclude?

Key Concepts

Null Hypothesis

The null hypothesis is the assumption that there is no significant difference in the distribution of applications among the card categories across all mailings. In other words, the proportions applying for Silver, Gold, and Platinum are the same in each mailing.

Chi-Square Test for Homogeneity

This test is used to determine whether different populations (in this case, mailings) have the same distribution of a categorical variable (card type applications). It compares observed frequencies to expected frequencies under the assumption that the distribution is homogeneous across groups.

Expected Frequencies

Expected frequencies are calculated based on the assumption that the null hypothesis is true. They are determined by multiplying the total for a given row and the total for a given column, then dividing by the overall total. This allows for a comparison between the observed counts and what would be expected if there were no differences across groups.

Chi-Square Statistic

The chi-square statistic is a measure of how much the observed frequencies deviate from the expected frequencies. It is calculated by summing the squares of the differences between observed and expected frequencies, divided by the expected frequency for each cell, providing a single metric to assess the goodness of fit.

Degrees of Freedom

Degrees of freedom in a chi-square test for homogeneity are calculated with the formula (number of rows - 1) multiplied by (number of columns - 1). This value reflects the number of independent pieces of information available for estimating variability in the observed data.

Critical Value

The critical value is the threshold obtained from the chi-square distribution corresponding to a specific significance level (alpha) and degrees of freedom. It is used to decide whether to reject the null hypothesis; if the calculated chi-square statistic exceeds this critical value, the null hypothesis is rejected.

Conclusion of Hypothesis Test

Based on the chi-square statistic and critical value, a conclusion is drawn regarding the null hypothesis. If the statistic is greater than the critical value, there is sufficient evidence to conclude that the distribution of applications differs significantly among the mailings; if not, the data does not provide enough evidence to reject the null hypothesis.

Transcript

00:01 For this problem, we need some information back from when we solved exercise 2 about bank cards.

00:07 In that problem, we had three categories of bank cards.

00:11 I believe it was gold, silver, and platinum.

00:16 And because we had three categories, we had two degrees of freedom, and we calculated a kai squared value of 12 .5.

00:24 So that's the information you need from exercise 2 in order to solve this problem.

00:29 Let's start in on this problem with part 8.

00:32 So in part a, if customers apply for all three cards, according to the historical proportions, how big on average would you expect the kai square distribution to be? so this is really asking us what is the mean of the kai square distribution? and in order to answer that, we need to know that the mean of any kai square distribution is just the degrees of freedom.

00:55 This is the expected value.

00:57 You could say the mean, but you could also say the expected value.

01:02 Of the distribution.

01:04 So it's just going to be equal to the degrees of freedom, which here is two.

01:08 So our mean of our kai square value is two.

01:12 B then asks us to compare that to the statistic that we calculated kai squared value was actually 12 .5.

01:24 Does that seem big or small in comparison to our mean? it does.

01:29 It seems large.

01:31 Our 12 .5 is a lot.

01:32 Bigger than our expected value.

01:35 So we're going to say it seems large in comparison.

01:40 Then it asks us what that means.

01:42 That means we have a sense that arch high square, this might be a weird event that we have witnessed.

01:50 It might be deviating from our expected value or deviating from our null hypothesis.

01:56 So we might be able to say there might be enough evidence to reject the null hypothesis.

02:07 And what do i mean when i say reject the null hypothesis? the null hypothesis is the statement that these data resemble the historical data.

02:24 So the null hypothesis is always kind of the boring hypothesis.

02:27 We have historical data on the percentages that these bank card applications fell into.

02:33 Did our new data exactly match the historical data? that would be the null hypothesis.

02:38 We have enough interesting kai squared information to maybe say no we don't match the old null hypothesis.

02:46 Here is how we check it.

02:48 So sort of b and c, we've kind of answered already...

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