i need to test the hypothesis that a proportion is greater than 0.93 using a confidence interval. i cannot figure out how to do this. the basic information that i think i need to use is below. the underlying variable follows a binomial distribution.
$$\hat p = 0.95, p = 0.93, n = 100, \sigma = 0.4$$
i have already tried constructing a confidence interval for a proportion using the Z distribution, but the upper bound is larger than 1, which seems wrong.
any advice for what i should do?
The problem of your exercise is that the data are NOT coherent.
Substitute the given $\sigma=0.4$ with its estimation $\hat{\sigma}=\sqrt{0.93\cdot0.07}\approx0.255$ and you will reach a nice solution.
You need also to fix a Type I error ($\alpha$)
Example: fixing $\alpha=2.5\%$ you get that the one-tail CI is
$$[0.93;0.93+\frac{1.96}{10}0.255]=[0.93;0.98]$$
You cannot reject your null hypothesis as $0.95$ is included in the interval