Question

Given below are the number of successes and sample size for a simple random sample from a population. xequals6​, nequals50​, 90​% level a. Determine the sample proportion. b. Decide whether using the​ one-proportion z-interval procedure is appropriate. c. If​ appropriate, use the​ one-proportion z-interval procedure to find the confidence interval at the specified confidence level. d. If​ appropriate, find the margin of error for the estimate of p and express the confidence interval in terms of the sample proportion and the margin of error.

Answer 1

a. Sample proportion ^p= 0.12

b. It is appropiate.

c. [0.0447;0.1953]

d. [^p ± d]

Explanation:

Hello!

Given the information I'll assume that the variable of study has a binomial distribution:

X~Bi(n;ρ)

The sample data:

n= 50

"Success" x= 6

Sample proportion ^p= x/n = 6/50 = 0.12

Now, your study variable has a binomial distribution, but remember that the Central Limit Theorem states that given a big enough sample size (usually n≥ 30) you can approximate the sample proportion distribution to normal.

Since the sample is 50 you can apply the approximation, your sample proportion will have the following distribution:

^p≈ N( p; [p(1 - p)]/n)

With E(^p)= p and V(^p)= [p(1 - p)]/n.

This allows you to estimate the population proportion per Confidence Interval using the Z-distribution:

[^p±
`Z_(1-\alpha /2)`*√(^p(1 - ^p)/n)]

Since you are estimating the value of p, you'll use the estimated standard deviation (i.e. with the sample proportion instead of the population proportion)

to calculate the interval.

At level 90% the interval is:

[0.12±1.64*√([0.12(1 - 0.12)]/50)]

[0.0447;0.1953]

The margin of error (d) of an interval is half its amplitude (a)

if a= Upper bond - Low bond

then d= (Upper bond - Low bond)/2

d= (0.1953-0.0447)/2

d= 0.0753

And since the interval structure is "estimator" -/+ "margin of error" you can write it as:

[^p ± d]

I hope you have a SUPER day!