Question

A random sample of 100 high school students was surveyed regarding their favorite subject. The following counts were obtained: Favorite Subject Number of Students English Math Science 30 Art/Music The researcher conducted a test to determine whether the proportion of students was equal for all four subjects. What is the value of the test statistic? O b. 25 OOOO d. -4 How many degrees of freedom does the chi-square test statistic for a goodness of fit have when there are 10 categories? a. 74 OOOO d. 62

Answer 1

Answer:

a) $\chi^2 = \frac{(25-25)^2}{25}+\frac{(30-25)^2}{25}+\frac{(30-25)^2}{25}+\frac{(15-25)^2}{25}=6$

b) $df=Categories-1=10-1=9$

Step-by-step explanation:

We assume the following info:

Favorite Subject         Number of students

English                                    25

Math                                        30

Science                                   30

Art/Music                                 15

Total                                        100

Previous concepts

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".

Part a

The system of hypothesis on this case are:

H0: There is no difference with the distribution proposed

H1: There is a difference with the distribution proposed

The level os 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 table given represent the observed values, we just need to calculate the expected values are 25 for each category.

And the calculations are given by:

$E_{English} =25$

$E_{Math} =25$

$E_{Science} =25$

$E_{Music} =25$

And now we can calculate the statistic:

$\chi^2 = \frac{(25-25)^2}{25}+\frac{(30-25)^2}{25}+\frac{(30-25)^2}{25}+\frac{(15-25)^2}{25}=6$

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

$df=Categories-1=4-1=3$

And we can calculate the p value given by:

$p_v = P(\chi^2_{3} >6)=0.112$

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

"=1-CHISQ.DIST(6,3,TRUE)"

Part b

For this case we have this formula:

$df=Categories-1=10-1=9$