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?

a)

b)

c)

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

The statistic to check the hypothesis is given by:

The observed values are:

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:

b) Compute the chi squared statistic.

And now we can calculate the statistic:

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

And we can calculate the p value given by:

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.

this is false ^.^ lol

$4034.40

the perimeter of the field is the sum of the lengths of the four sides:

47 yd + 158 yd + 47 yd + 158 yd = 410 yd

the cost of 3 ft (1 yd) of fence is 3×$3.28 = $9.84.

so the cost of jake's wood perimeter fance is

(410 yd) × ($9.84 /yd) = $4034.40