# Hypothesis test for two sample proportions?

I need help with this question. I have the formula and all, but i always get the wrong answer

In a sample of 200 men, 130 said they used seat belts. In a sample of 300 women, 63 said they used seat belts. At significance level of 0.01, test the claim that the proportions of men that use seat belts is higher than the proportions of women that do

### 1 Answer

- IndicaLv 710 years agoFavorite Answer
Are you sure you have the data right ? 163 would be more realistic. Anyway I've used the given data to illustrate the method. You can always change the numbers.

pm = rv : proportion of men in population using seat belts

pw = rv : proportion of women in population using seat belts

H0 : pm=pw

H1 : pm>pw

qm = rv : proportion in 200 sample

qw = rv : proportion in 300 sample

Step #1 : Assume H0 is true. Then the two samples have been drawn from the same population so we pool them and use the average over both samples p = (130+63)/(200+300) = 0.386 as our best estimate of pm and pw. So under H0, pm=p and pw=p.

Step #2 : The test-statistic is t = qm−qw. We want to test whether or not the value of this for the given samples is in the critical region. But first we develop the pdf for t.

For each of the two samples we know from the distribution of sample proportions that

qm ~ N( 0.386, 0.386*0.614/200 ) and qw ~ N( 0.386, 0.386*0.614/300 )

∴ t = qm−qw ~ N( 0, 0.386*0.614(1/200+1/300) ) = N( 0, 1.975*10‾³ ) So we have it.

The normalised deviate is z = (t−0) / √1.975*10‾³

Step #3 : The value of t from the samples is t = 0.65−0.21 = 0.44 (!!) and the corresponding value of z is 0.44 / √1.975*10‾³ = 9.9

Step #4 : Since H1 is pm>pw we do a one-tailed test. The critical region is defined by P(z≥zcrit)=0.01 so from N(0,1) tables zcrit=2.33.

Step #5 : Clearly 9.9>2.33 so H0 is rejected and H1 is accepted.

Merry Xmas

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