Answer
a. The claim cannot be supported.
b. The claim cannot be supported.
c. No
Work Step by Step
a. We want to see if the mean is different than $\frac{40}{45}=.8889$, so we know $H_0: \mu = .8889$ and $H_1: \mu \ne .8889$. We now use a Ti-84 calculator. To solve, go to "Stat," then "Tests," and then select "Stats." Doing this, we find that the critical value and use a z-table calculator to find that p is 0.5720. This is more than .05, so the claim is not supported.
b. We use a ti-84 to find the confidence interval. To do this, we press “Stat” and then select “Tests.” Next, we select “2-PropZInt” to conduct the test. Doing this, we find that the confidence interval is $-.0798 \lt \mu_1-\mu_2\lt .149$. This goes through 0, so we see that there is not a significant difference.
c. The results show that it does not have an impact on infection rate.
