Question

A child psychologist believe that controlled physical outbursts of anger (like punching a pillow) may improve the mood of young boys with emotional impairment. He believes that the proportion of boys who benefit from the treatment is greater than the proportion of girls. A random sample from each population receives counseling in the treatment and is asked about their mood after an episode (x is the number of test subjects that reported an improvement in mood). The results of the study are summarized in the table below. Boys Girls n1 = 200 X1 = 35 n2 = 210 X2 = 28 Blank #1: Find the two sample proportions, reporting each as a decimal, rounded to two decimal places. Do the initial (untested) findings show what the researcher expected? Enter your answer as a list of 3 items: P1, P2, yes or no. Do not add any extra spaces and pay attention to rounding. For example, a valid answer may look like .52,14 yes or 25,.50,no. Untested, you compared them but you could have done that before you took stats class:) Comparing the sample proportions does not build in any uncertainty or variation into the estimates; thus we use inference procedures such as confidence intervals and significance tests to do so.

Answer 1

solution:

the given information as follows:

sample details for boys:

ni 200
,

proportion of boys who report an improvement = = 35/200 = 0.175

P1

for sample 2:

n2 = 210, X2 = 28

proportion of female who report an improvement = = 28/210 = 0.1333

d

value of the pooled proportion is computed as $\bar p=\frac{X_1+X_2}{n_1+n_2}=\frac{35+28}{200+210}=0.1537$

we have to test is the proportion of boys is greater than proportion of girls

so, null and alternative hypothesis:

Ho: P1 = P2

it is a right tailed test:

test statistics:

$z=\frac{\hat p_1-\hat p_2}{\sqrt{\bar p(1-\bar p)\left ( \frac{1}{n_1}+\frac{1}{n_2} \right )}}=\frac{0.175-0.1333}{\sqrt{0.1537(1-0.1537)\left ( \frac{1}{200}+\frac{1}{210} \right )}}=1.17$

p value = 1 - value of z to the left of 1.17 from z table = 1 - 0.8789 = 0.1211

p value 0.1211 0 1.17

this p value in tha area of the right tailed formed by z = 1.17.

let significance level = $\alpha$ = 0.05

since p value = 0.1211 > 0.05, so do not reject the null hypothesis H0

conclusion;

evidence does not favour that the proportionof boys whoo benefit from the treatment is greater than the proportion of girls.