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 0.2 significance level.

The null and alternative hypothesis would be:

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

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:μM=μFH0:μM=μF
H1:μM≠μFH1:μM≠μF

The test is:

two-tailed

left-tailed

right-tailed

Based on a sample of 40 men, 30% owned cats
Based on a sample of 20 women, 50% 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

Answer)

Null hypothesis usually have three types of claim.

1) >= (greater than or equal to )

2) <= (less than or equal to)

3) = (equal to )

Alternate hypothesis Ha have

1) < (less than (left tailed ))

2) > (greater than (right tailed))

3) not equal to (two tailed))

Both are exactly opposite of each other and in some cases we just use = sign for null hypothesis.

1)

H0:pM=pFH0:pM=pF
H1:pM≠pFH1:pM≠pF

The test is two tailed.

N1 = 40, P1 = 0.3.

N2 = 20, P2 = 0.5.

First we need to check the conditions of normality that is if n1p1 and n1*(1-p1) and n2*p2 and n2*(1-p2) all are greater than equal to 5 or not

N1*p1 = 12

N1*(1-p1) = 28

N2*p2 = 10

N2*(1-p2) = 10

All 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 = [(p1*n1)+(p2*n2)]/[n1+n2]

After substitution

Test statistics z = -1.52

From z table, P(Z<-1.52) = 0.0643

Sjnce test is two tailed,

So, P-Value = 0.0643*2 = 0.1286

As the obtained P-Value is less than the given significance 0.2.

Reject the null hypothesis.