Question

At a major credit card​ bank, the percentages of people who historically apply for the​ Silver, Gold and Platinum cards are 60 %​, 30 % and 10 % respectively. In a recent sample of customers responding to a​ promotion, of 200​ customers, 103 applied for​ Silver, 68 for Gold and 29 for Platinum. Is there evidence to suggest that the percentages for this promotion may be different from the historical​ proportions? ​a) What is the expected number of customers applying for each type of card in this sample if the historical proportions are still​ true? ​b) Compute the chi squared statistic. ​c) How many degrees of freedom does the chi squared statistic​ have?

Answer 1

Answer:

a) $E_{Gold} =0.3*200=60$

$E_{Silver} =0.6*200=120$

$E_{Platinum} =0.1*200=20$

b) $\chi^2 = \frac{(68-60)^2}{60}+\frac{(103-120)^2}{120}+\frac{(29-20)^2}{20} =7.525$

c) $df=Categories-1=3-1=2$

Step-by-step explanation:

A chi-square goodness of fit test "determines if a sample data matches a population".

A chi-square test for independence "compares two variables in a contingency table to see if they are related. In a more general sense, it tests to see whether distributions of categorical variables differ from each another".

Assume the following dataset:

Silver, Gold and Platinum cards are 60 %​, 30 % and 10 %

We need to conduct a chi square test in order to check the following hypothesis:

H0: There is no difference with the proportions claimed

H1: There is a difference with the proportions claimed

The level of significance assumed for this case is $\alpha=0.05$

The statistic to check the hypothesis is given by:

$\chi^2 =\sum_{i=1}^n \frac{(O_i -E_i)^2}{E_i}$

The observed values are:

$O_{Gold}=68$

$O_{Silver}=103$

$O_{Platinum}=29$

a) What is the expected number of customers applying for each type of card in this sample if the historical proportions are still​ true?

The expected values are given by:

$E_{Gold} =0.3*200=60$

$E_{Silver} =0.6*200=120$

$E_{Platinum} =0.1*200=20$

b) Compute the chi squared statistic. ​

And now we can calculate the statistic:

$\chi^2 = \frac{(68-60)^2}{60}+\frac{(103-120)^2}{120}+\frac{(29-20)^2}{20} =7.525$

Now we can calculate the degrees of freedom for the statistic given by:

$df=Categories-1=3-1=2$

And we can calculate the p value given by:

$p_v = P(\chi^2_{2} >7.525)=0.0232$

And we can find the p value using the following excel code:

"=1-CHISQ.DIST(7.525,2,TRUE)"

Since the p value is lower than the significance level we reject the null hypothesis at 5% of significance, and we can conclude that we have significant differences from the % assumed for each category.