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 ni = 200 X1 = 35 n2 = 210 x2 = 28 Let P1 represent the proportion of boys who benefit from treatment and P2 be the proportion of girls. To explore the suspicion above, conduct a significance test using the hypotheses: Blank #1: Find the sample proportion for boys 1. Round to two decimal places. Blank #2: Find the sample proportion for girls, P 2. Round to two decimal places. Blank #3: Do the initial (untested) fidnings show what the researcher expected? (Yes or No). Blank #4: Determine the correct alternative hypothesis by filling in the blank with <, > or not equals. P1 = P2 Pi -P2 Blank #5: The test statistic of this test is z = 1.17 with corresponding p-value of p = 0.1211. If you were to draw the distribution of the sample proportions under the assumption of the null hypothesis, this p-value is the area in the

Answer 1

Given

Boys	Girls
n1 = 200	n2 = 210
x1 = 35	x2 = 28

We wish to test that weather proportion of boys benifited from the treatment is greater than proportion of girls or not.

Blank #1

Sample proportion for boys is given by

$\hat{p1}$ = x1/n1 = 35/200 = 0.175

Thus $\hat{p1}$ = 0.175 $\approx$ 0.18           { rounding to two decimals }

Blank #2

Sample proportion for girls is given by

$\hat{p2}$ = x2/n2 = 28/210 =   0.1333

Thus $\hat{p2}$ = 0.1333 $\approx$ 0.13          { rounding to two decimals }

Blank #3

Yes

Since sample proportin for boys $\hat{p1}$ = 0.18 is greater than Sample proportion for girls $\hat{p2}$ = 0.13

i.e $\hat{p1}$ > $\hat{p2}$

So answer is Yes , initial finding shows what the reaserch expected i.e proportion of boys benifited from the treatment is greater than proportion of girls

Blank #4

To Test :

H0 : p1 = p2      { proportion of boys benifited may be same to that of the proportion of girls }

H1 : p1 > p2      { proportion of boys benifited from the treatment is greater than proportion of girls }

hence correct option is   " > "

null hypothesis H0 : p1 = p2

alternative hypothesis H1 : p1 > p2

Blank #5

Test Statistics Value z = 1.17

P-Value = 0.1211

Since this is one-tail test and alternative hypothesis is of " > " type , so p-value is Right-tail area .

Hence correct option is

The P-Value is the area in Right tail form by z=1.17

Blank #6

We reject null hypothesis if corresponding P-value is less than given level of significance

We are given $\alpha$ = 0.05 level of significance .

As P-Value = 0.1211 > 0.05   i.e P-value is greater that $\alpha$ , we fail to reject null hypothesis at 0.05 level of significance

Decision : - Do not reject H0

Blank #7

Since we fail to reject null hypothesis , i.e we don't have enough evidence against null hypothesis. So we cannot conclude that proportion of boys benifited is greater than proportion of girls

So correct option is

Evidence dose not favor that the proportion of boys benifited from the treatment is greater than proportion of girls

Blank #8

No

Since , in initial untested finding , i.e untested answer was that the proportion of boys benifited is greater than proportion of girls , but final evidence after testing we draw is different from untested answer .

Blank # 1 :- 0.18

Blank # 2 :- 0.13

Blank # 3 :- Yes

Blank # 4 :- " > "

Blank # 5 :- Right

Blank # 6 :- Do not reject H0

Blank # 7 :- Does not favor

Blank # 8 :- No