Answer:
a)
![E_(Gold) =0.3*200=60](https://img.qammunity.org/2020/formulas/mathematics/college/3p0e74ud3q2fi2lv730qb9841aoztobe9m.png)
![E_(Silver) =0.6*200=120](https://img.qammunity.org/2020/formulas/mathematics/college/lri4pwdtd9duvmvgai0a5xg1l9wyrowfa7.png)
![E_(Platinum) =0.1*200=20](https://img.qammunity.org/2020/formulas/mathematics/college/qxxtz5h2o80pxow2k5gox9yvbylp62rece.png)
b)
![\chi^2 = ((68-60)^2)/(60)+((103-120)^2)/(120)+((29-20)^2)/(20) =7.525](https://img.qammunity.org/2020/formulas/mathematics/college/6z3zxt887mze4hk3rivph1h4t4p0xf5uda.png)
c)
![df=Categories-1=3-1=2](https://img.qammunity.org/2020/formulas/mathematics/college/x55bi6sh040714sjvkvioy7vus8g7dlmwn.png)
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](https://img.qammunity.org/2020/formulas/mathematics/college/o3op132eurfz836qnoznjuckj0omh3ecx4.png)
The statistic to check the hypothesis is given by:
![\chi^2 =\sum_(i=1)^n ((O_i -E_i)^2)/(E_i)](https://img.qammunity.org/2020/formulas/mathematics/college/m8fldj4qkykpo49zkyl0byr8hqgmttkcm7.png)
The observed values are:
![O_(Gold)=68](https://img.qammunity.org/2020/formulas/mathematics/college/n5geaf61qhx8ji3p39wm1cfcpz4kh312gr.png)
![O_(Silver)=103](https://img.qammunity.org/2020/formulas/mathematics/college/hvc56x81a7gj8zhrg9fdld9oik12u08dj7.png)
![O_(Platinum)=29](https://img.qammunity.org/2020/formulas/mathematics/college/6w4e9nn02tzyatlzlonad5zd60aw88c7vq.png)
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](https://img.qammunity.org/2020/formulas/mathematics/college/3p0e74ud3q2fi2lv730qb9841aoztobe9m.png)
![E_(Silver) =0.6*200=120](https://img.qammunity.org/2020/formulas/mathematics/college/lri4pwdtd9duvmvgai0a5xg1l9wyrowfa7.png)
![E_(Platinum) =0.1*200=20](https://img.qammunity.org/2020/formulas/mathematics/college/qxxtz5h2o80pxow2k5gox9yvbylp62rece.png)
b) Compute the chi squared statistic.
And now we can calculate the statistic:
![\chi^2 = ((68-60)^2)/(60)+((103-120)^2)/(120)+((29-20)^2)/(20) =7.525](https://img.qammunity.org/2020/formulas/mathematics/college/6z3zxt887mze4hk3rivph1h4t4p0xf5uda.png)
Now we can calculate the degrees of freedom for the statistic given by:
![df=Categories-1=3-1=2](https://img.qammunity.org/2020/formulas/mathematics/college/x55bi6sh040714sjvkvioy7vus8g7dlmwn.png)
And we can calculate the p value given by:
![p_v = P(\chi^2_(2) >7.525)=0.0232](https://img.qammunity.org/2020/formulas/mathematics/college/87pzgfj0dl6w5kuuv46u14159xliyjo31o.png)
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.