Q. 29 - The Practice of Statistics for AP

Question

Refer to Exercise 27. Do the data provide convincing evidence of a difference in the distributions of sports goals for male and female undergraduates at the university? 

(a) State appropriate null and alternative hypotheses for a significance test to help answer this question. 

(b) Calculate the expected counts. Show your work. 

(c) Calculate the chi-square statistic. Show your work 

From exercise $27$

Step-by-Step Solution

Answer

(a) $H_0$There is no difference between the distribution of sports goals for females and the distribution of sports goals for males.

 $H_a$: There is a difference between the distribution of sports goals  for females and the distribution of sports goals for males.

(b) 

(c) The value of the chi-square statistic is 24.8978.

Part (a) Step 1: Given information

The data is 

Part (a) Step 2: Explanation

The null hypothesis states that there is no association between the variables:

$H_0$ : There is no difference between the distribution of sports goals for females and the distribution of sports goals for males.

The alternative hypothesis states that there is an association between the variables: 

$H_a$: There is a difference between the distribution of sports goals for females and the distribution of sports goals for males.

Part (b) Step 1: Given information

The data given is 

Part (b) Step 2: Explanation

From the given 

$n_1=67$

$n_2=67$

$$\begin{gathered} n=67+67 \\ =134 \end{gathered}$$

The expected count is the row total multiplied by the sample size divided by the total sample size $n=134$

Part (c) Step 1: Given information

The given data is

Part (c) Step 2: Explanation

Observed counts

Expected counts

The chi-square statistic is the sum of squared deviations (between observed and expected counts) divided by the expected count:

$$\begin{gathered} {\chi }^2=\frac{(14-22.5{)}^2}{22.5}+\frac{(7-12.5{)}^2}{12.5}+\frac{(21-13{)}^2}{13}+\frac{(25-19{)}^2}{19}+\frac{(31-22.5{)}^2}{22.5}+\frac{(18-12.5{)}^2}{12.5}+\frac{(5-13{)}^2}{13}+\frac{(13-19{)}^2}{19} \\ \approx 24.8978 \end{gathered}$$