Question

If a simple random sample of 200 provides 50 'yes' responses, the 89% confidence interval for the population proportion is I need help to understand how I can find Z value from the z table!!! Please help!!!

Answer 1

We know that 89% confidence interval for population proportion is given by: $(p\pm Z_{0.11/2}\sqrt{\frac{p(1-p)}{n}})$

Now, p=50/200 = 0.25

$(0.25\pm Z_{0.055}\sqrt{\frac{0.25(1-0.25)}{200}})$

Now, $Z_{0.055}$ is the value 'a' such that $P(Z>a)=0.055$

So, using Normal probability tables, we get $Z_{0.055}=1.60$

So, the confidence interval becomes, $(0.25\pm (1.60)\sqrt{\frac{0.25(1-0.25)}{200}})$

$(0.25\pm 0.0490)$

$(0.2010,0.2990)$