Question

Test whether males are less likely than females to support a ballot initiative, if 23% of a random sample of 52 males plan to vote yes on the initiative and 31% of a random sample of 52 females plan to vote yes.

Use subscripts 1 for males who plan to vote yes and 2 for females who plan to vote yes.

(a)Find the relevant sample proportions in each group and the pooled proportion.

Round your answers to three decimal places, if necessary.

P-hat 1=

P-hat 2=

P-hat=

(b) State the null and alternative hypotheses.

(c) Give the test statistic and the p-value.

Round your answer for the test statistic to two decimal places and your answer for the p-value to three decimal places.

Answer 1

Answer)

A)

Pooled proportion = (n1*p1 + n2*p2)/(n1+n2)

N1 = N2 = 52

P1 = 0.23 and P2 = 0.31

After substitution

Pooled proportion = 0.27

P-hat 1 = 0.23

P-hat 2 = 0.31

P-hat = 0.27

B)

Null hypothesis Ho : P 1 = P2

Alternate hypothesis Ha : P1<P2

C)

First we need to check the conditions of normality that is if n*p and n*(1-p) both are greater than 5 or not

N1*p1 = 12

N2*p2 = 16

Both the conditions are met so we can use standard normal z table to conduct the test

Test statistics z = (P1-P2)/standard error

Standard error = √{p*(1-p)}*√{(1/n1)+(1/n2)}

P = pooled proportion

Z = -0.92

From z table, P(Z<-0.92) = 0.1788

P-value = 0.1788