I need the work shown for solving the problems to compare it with my own work. I

Question

I need the work shown for solving the problems to compare it with my own work. I also use a TI-84 Plus calculator if a formula is needed, Thanks

15) Given a sample standard deviation of 50 and a sample size of 20, calculate the standard error of the mean. a 11.18 b) 2.5 c) 50 d) 20 16) What is the most conservative probability of success when creating confidence intervals about a sample proportion. b) .1 d) changes with sample size 17) Create a 99% confidence interval about the sample proportion given that the proportion is .42 and the sample size is 500 a) .42 b) .0568 36-477 d) 4-44 18) Find the sample size needed to stay within 2 units of the sample mean given the sample standared deviation is 12 and you are going to create a 90% confidence interval 98 b) 49 c) 9.87 d) 24

Explanation / Answer

Solution:-

15).

Given, Sample standard eviation = 50, and sample size = 20

Standard error of sample mean = / sqrt(n) = 50 / sqrt(20) = 11.18

16). For the following procedures, the assumption is that both np 10 and n(1p) 10.

If p is unknown, we use pˆ as an estimate of p.

A confidence interval for a populaton proportion is constructed by taking the point estimate (pˆ) plus and minus the margin of error.

The margin of error is computed by multiplying a z multiplier by the standard error, SE(pˆ).

pˆ ± z * ((p^ * (1p^)) / n)

Thus, 0.5 is the most conservative probability of success, as p and q(=1-p) both are 0.5.

17).

Formula:

Confidence Interval = p ± Z/2 × [(p×q)/n] , (x, n-x 5)

Where, p = x/n => p = 0.42

q = 1-p => q = 1 - 0.42

= 1 - (Confidence Level/100) =>   = 0.01

x = Frequency => p = x/n => 0.42 = x/500 => x = 500*0.42 = 210

n = Sample Size = 500

Z/2 = Z-table value = 2.575

Confidence Interval for Proportion, 0.363 < p < 0.477

18).

Formula, n = [ ( Z/2 * ) / E ]2

n = [ (1.645 * 12) / 2 ]2

      n = 97.4169

or n = 98

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