Answer:
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:
: p2=p1
: 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:
where
- p(s) is the sample proportion of high school teachers who were single (
) - p is the proportion of elementary teachers who were single (
)
- N is the sample size (180)
Using the numbers, we get
≈ 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)
keeping in mind that when the logarithm base is omitted, the base 10 is assumed.
![\textit{exponential form of a logarithm} \\\\ \log_a(b)=y \qquad \implies \qquad a^y= b \\\\[-0.35em] ~\dotfill\\\\ \log(x)=2\implies \log_{10}(x)=2\implies 10^2=x\implies 100=x](https://tex.z-dn.net/?f=%5Ctextit%7Bexponential%20form%20of%20a%20logarithm%7D%20%5C%5C%5C%5C%20%5Clog_a%28b%29%3Dy%20%5Cqquad%20%5Cimplies%20%5Cqquad%20a%5Ey%3D%20b%20%5C%5C%5C%5C%5B-0.35em%5D%20~%5Cdotfill%5C%5C%5C%5C%20%5Clog%28x%29%3D2%5Cimplies%20%5Clog_%7B10%7D%28x%29%3D2%5Cimplies%2010%5E2%3Dx%5Cimplies%20100%3Dx)
Answer: 0.72
Step-by-step explanation:
Hope that helps :)
I don’t know but I need the Points sorry for you