Question

Test the claim that the proportion of men who own cats is significantly different than the proportion of women who own cats at the 0.02 significance level.

The null and alternative hypothesis would be:

H0:μM=μFH0:μM=μF
H1:μM≠μFH1:μM≠μF

H0:pM=pFH0:pM=pF
H1:pM≠pFH1:pM≠pF

H0:pM=pFH0:pM=pF
H1:pM<pFH1:pM<pF

H0:μM=μFH0:μM=μF
H1:μM<μFH1:μM<μF

H0:μM=μFH0:μM=μF
H1:μM>μFH1:μM>μF

H0:pM=pFH0:pM=pF
H1:pM>pFH1:pM>pF

The test is:

right-tailed

two-tailed

left-tailed

Based on a sample of 60 men, 45% owned cats
Based on a sample of 60 women, 60% owned cats

The test statistic is:  (to 2 decimals)

The p-value is:  (to 2 decimals)

Answer 1

Solution :

Given that,

n₁ = 60

Point estimate = sample proportion = ₁ = 0.45

n₂ = 60

Point estimate = sample proportion = ₂ = 0.60

The value of the pooled proportion is computed as,

= (n_{11 +} n₂₂  ) / ( n₁ + n₂ )

= (27 + 36 ) / (60 + 60 )

= 0.525

1 - = 0.475

Level of significance = = 0.02

The null and alternative hypothesis is,

H0:μM=μF

H1:μM≠μF

This a two-tailed test.

Test statistics

z = (₁ - ₂ ) / *(1-) ( 1/n₁ + 1/n₂ )

= (0.45 - 0.60 ) / (0.525 * 0.475 ) (1/60 + 1/60 )

= -1.65

P-value

= 2* P(Z <z )

= 2* P(Z < -1.65)

= 2*0.0495

= 0.099

= 0.10

The p-value is p = 0.10, and since p = 0.10 > 0.02, it is concluded that the null hypothesis is fail to rejected.