Question

A random sample of 115 observations results in 46 successes. Use Table 1.
a. Construct a 90% confidence interval for the population proportion of successes. (Round intermediate calculations to 4 decimal places, "z" value to 2 decimal places, and final answers to 3 decimal places.)
Confidence interval to
b. Construct a 90% confidence interval for the population proportion of failures. (Round intermediate calculations to 4 decimal places, "z" value to 2 decimal places, and final answers to 3 decimal places.)
Confidence interval to

Answer 1

Option a) 90% confidence interval for the population proportion of successes = 0.402

Option b) 90% confidence interval for the population proportion of failures = 0.598

According to the question given,

Critical values of Standard deviations are to be used to construct the proportions for the confidence interval.

The formula of a confidence interval that we will be using for the solution will be,

p = $z_{a} * \sqrt{\frac{p(1-p)}{n} }$

In the above expression,

p = observed population.

n = sample size

$z_{a}$ = critical value of confidence interval

Now the given observations in question = 115

That result in success = 46

The resultant sample proportion = $\frac{46}{115}$

⇒ 0.40 and it is for both success and failure

Here we can see that the answer for both parts will be identical.

The critical value of z at 90% confidence = 1.64

a) Lower bound = $0.4 - 1.64 * \sqrt{\frac{0.4(1-0.4)}{90} }$

⇒ 0.402

b) Upper bound = $0.4 + 1.64 * \sqrt{\frac{0.4(1-0.4)}{90} }$

⇒ 0.598

Confidence interval = 0.402 to 0.598

Therefore,

Option a) 90% confidence interval for the population proportion of successes = 0.402

Option b) 90% confidence interval for the population proportion of failures = 0.598

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