Question

Assume that you plan to use a significance level of ΅= 0.05 to test the claim that p1= p2. Use the given sample sizesand numbers of successes to find the z test statistic for the hypothesis test.8)8)Information about movie ticket sales was printed in a movie magazine. Out of fifty PG-ratedmovies, 41% had ticket sales in excess of $3,000,000. Out of thirty-five R-rated movies, 17%grossed over $3,000,000.A)z =3.763B)z =2.352C)z =7.291D)z =4.704

Assume that you plan to use a significance level of ΅= 0.05 to test the claim that p1= p2, Use the given sample sizesand numbers of successes to find the P-value for the hypothesis test.9)9)n1= 100n2= 140x1= 41x2= 35A)0.0512B)0.4211C)0.0021D)0.0086

Use the traditional method to test the given hypothesis. Assume that the samples are independent and that they havebeen randomly selected10)10)In a random sample of 360 women, 65% favored stricter gun control laws. In a random sample of220 men, 60% favored stricter gun control laws. Test the claim that the proportion of womenfavoring stricter gun control is higher than the proportion of men favoring stricter gun control.Use a significance level of 0.05. What is the test statistic, critical value, and interpretation?A)Test stat: z= 1.21, Crit Value z= 1.645, Reject the nullB)Test stat: z= 2.42, Crit Value z= 1.645, Fail to reject the nullC)Test stat: z= 2.42, Crit Value z= 1.645, Fail to reject the nullD)Test stat: z= 1.21, Crit Value z= 1.645, Fail to reject the null

Answer 1

(8) Here, n1 =50, p1=41% = 41/100 = 0.41

n2 =35, p2=17% = 17/100 = 0.17

To test the claim p1=p2, we compute the following

P=(n1p1+n2p2)/(n1+n2) = (50x0.41+35x0.17)/(50+35)

=26.45/85 = 0.3111

Q=1-P = 1-0.3111=0.6889

SE(p1-p2) = SQRT[PQ{(1/n1)+(1/n2)}]

= SQRT[0.3111x0.6889{(1/50)+(1/35)}]

= SQRT[0.3111x0.6889x0.0485]

=0.1019

The test statistic Z = (p1-p2)/SE(p1-p2)

=(0.41-0.17)/0.1019

=0.24/0.1019

=2.3552

Answer, B)Z=2.352

(9) Here, n1 =100, x1=41 , n2 =140, x2=35

p1=x1/n1 = 41/100 = 0.41

p2 = x2/n2 =35/140 =0.25

To test the claim p1=p2, we compute the following

P=(x1+x2)/(n1+n2) = (41+35)/(100+140)

=76/240 = 0.3166

Q=1-P = 1-0.3166 = 0.6834

SE(p1-p2) = SQRT[PQ{(1/n1)+(1/n2)}]

= SQRT[0.3166x0.6834{(1/100)+(1/140)}]

= SQRT[0.3166x0.6834x0.0171]

=0.0608

The test statistic Z = (p1-p2)/SE(p1-p2)

=(0.41-0.25)/0.0608

=0.16/0.0608

=2.63

For two tailed test, the P value = P(Z>2.63) /2 = [1-P(Z<=2.63)]/2 = [1-0.9957]/2 = 0.0043/2 = 0.00215

Answer, C)0.0021

(10) ) Here, for women, n1 =360, p1=65% = 65/100 = 0.65

For men, n2 =220, p2=60% = 60/100 = 0.6

To test the claim p1>p2, we frame null hypothesis H0:p1=p2 against alternative hypothesis H1:p1>p2

It is one-tailed and right tailed test and we compute the following:

P=(n1p1+n2p2)/(n1+n2) = (360x0.65+220x0.6)/(360+220)

=0.631

Q=1-P = 1-0.631=0.369

SE(p1-p2) = SQRT[PQ{(1/n1)+(1/n2)}]

= SQRT[0.631x0.369{(1/360)+(1/220)}]

=0.0412

The test statistic Z = (p1-p2)/SE(p1-p2)

=(0.65-0.6)/0.0412

=1.2135

The level of significance alfa = 0.05

The critical value is 1.645

Since, the computed value 1.2135 is more than the critical value 1.645, then null H0 is rejected.

Answer, A)Test stat: z= 1.21, Crit Value z= 1.645, Reject the null