Please help!!! Due today!! Picture of question + example. Please show work!!!
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Answer:
The critical value that corresponds to a confidence level of 99% is, 2.58.
Step-by-step explanation:
Consider a random variable <em>X</em> that follows a Binomial distribution with parameters, sample size <em>n </em>and probability of success <em>p</em>.
It is provided that the distribution of proportion of random variable <em>X, </em>, can be approximated by the Normal distribution.
The mean of the distribution of proportion is,
The standard deviation of the distribution of proportion is, .
Then the confidence interval for the population proportion <em>p</em> is:
The confidence level is 99%.
The significance level is:
Compute the critical value as follows:
That is:
Use the <em>z</em>-table for the <em>z-</em>value.
For <em>z</em> = 2.58 the P (Z < z) = 0.995.
And for <em>z</em> = -2.58 the P (Z > z) = 0.005.
Thus, the critical value is, 2.58.
