Question

Test the claim that the proportion of men who own cats is smaller than the proportion of women who own cats at the .10 significance level.

The null and alternative hypothesis would be:

H0:μM=μF________________
H1:μM<μF

H0:pM=pF_____________
H1:pM<pF

H0:pM=pF__________________
H1:pM≠pF

H0:μM=μF__________________
H1:μM>μF

H0:pM=pF______________
H1:pM>pF

H0:μM=μF ________________
H1:μM≠μF

The test is:

right-tailed___________

two-tailed_____________

left-tailed____________

Based on a sample of 20 men, 45% owned cats
Based on a sample of 20 women, 70% owned cats

The test statistic is: _______________(to 2 decimals)

The p-value is: ______________(to 2 decimals)

Based on this we:

• _____________Fail to reject the null hypothesis
• ______________Reject the null hypothesis

Answer 1

p_M : Proportion of men who own cats

p_F :  Proportion of women who own cats

claim : proportion of men who own cats is smaller than the proportion of women who own cats at the

The null and alternative hypothesis would be:

H0 :p_M = p_F
H1: p_M < p_F

The test is left tailed

Sample size for Men : n₁ : 20

Sample proportion of men who own cats : $\widehat{p}_{1}$ = 45/100 =0.45

Sample size for women : n₂ : 20

Sample proportion of men who own cats : $\widehat{p}_{2}$ = 70/100 =0.70

$\\Test\; \; Statistic : Z_{stat}=\frac{\widehat{p_{1}}-\widehat{p_{2}}}{\sqrt{\frac{\widehat{p_{1}}(1-\widehat{p_{1}})}{n_{1}}+ \frac{\widehat{p_{2}}(1-\widehat{p_{2}})}{n_{2}}}}\\\\\\ =\frac{0.45-0.7}{\sqrt{\frac{0.45(1-0.45)}{20}+ \frac{0.7(1-0.7)}{20}}}=\frac{-0.25} {0.1512}=-1.6529$

The test statistic : -1.65

For left tailed test:

Level of significance : $\alpha$ =0.10

The p-value is : 0.05

As P-Value i.e. is less than Level of significance i.e (P-value:0.0492 < 0.10:Level of significance); Reject Null Hypothesis

Based on this we

Reject the null hypothesis