By using hypothesis test at α = 0.01, we cannot conclude that the proportion of high school teachers who were single greater than the proportion of elementary teachers who were single

Step-by-step explanation:

let p1 be the proportion of elementary teachers who were single

let p2 be the proportion of high school teachers who were single

Then, the null and alternative hypotheses are:

$H_{0}$ : p2=p1

$H_{a}$ : p2>p1

We need to calculate the test statistic of the sample proportion for elementary teachers who were single.

It can be calculated as follows:

$\frac{p(s)-p}{\sqrt{\frac{p*(1-p)}{N} } }$ where

p(s) is the sample proportion of high school teachers who were single ( $\frac{36}{180} =0.2$ )
p is the proportion of elementary teachers who were single ( $\frac{15}{100} =0.15$ )

N is the sample size (180)

Using the numbers, we get

$\frac{0.2-0.15}{\sqrt{\frac{0.15*0.85}{180} } }$ ≈ 1.88

Using z-table, corresponding P-Value is ≈0.03

Since 0.03>0.01 we fail to reject the null hypothesis. (The result is not significant at α = 0.01)

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3 years ago