A sample of 9 production managers with over 15 years of experience has an average salary of $71,000 and a sample standard deviation of $18,000. Assuming that s = 18,000 is a reasonable estimate for f$sigma f$ what sample size would be needed to ensure that we could estimate the true mean salary of all production managers with more than 15 years experience within $4200 if we wish to be 95% confident? Place your answer, as a whole number, in the blank. Do not use a dollar sign, a comma, or any other stray mark. For examples, 34 would be a legitimate entry.

You want to estimate the average salary of production managers with more than 15 years of experience, for this you will use a 95% confidence interval with a maximum estimation error of 4200.

[tex]\bar X = 71000\\\\\alpha = 0.05\\\\Z_{\frac{\alpha}{2}}=1.95996\\\\\sigma = 18000\\\\\epsilon=4200[/tex]

The expression for the calculation of the sample size is given by:

[tex]n= (\frac{\sigma Z_{\frac{\alpha}{2}}}{\epsilon})^2=(\frac{18000*1.95996}{4200})^2=70.5571[/tex]

astha8579 astha8579 · Answer 2 · 2022-02-09T09:53:12+01:00

You can use the margin of error formula to get the needed sample size.

The needed sample size for the given condition is 71

What is the margin of error for large samples?

Suppose that we have:

Sample size n > 30
Sample standard deviation = [tex]s[/tex]
Population standard deviation = [tex]\sigma[/tex]
Level of significance = [tex]\alpha[/tex]

Then the margin of error(MOE) is obtained as

Case 1: Population standard deviation is known

[tex]MOE = Z_{\alpha /2}\dfrac{\sigma}{\sqrt{n}}[/tex]

Case 2: Population standard deviation is unknown.

[tex]MOE = Z_{\alpha /2}\dfrac{s}{\sqrt{n}}[/tex]

where [tex]Z_{\alpha/2}[/tex] is critical value of the test statistic at level of significance [tex]\alpha[/tex]

Using the above formula, we get the sample size needed as

Since the limit of error is $4200, thus,

MOE = $4200

The level of significance is [tex]\alpha[/tex] = 100 - 95% = 5% = 0.05

Critical value [tex]Z_{\alpha/2}[/tex] at 0.05 level of significance for two tailed test is 1.96

The standard deviation estimate(sample standard deviation) is s= $18000, thus, we have:

[tex]MOE = Z_{\alpha /2}\dfrac{s}{\sqrt{n}}[/tex]

[tex]4200 = 1.96 \times \dfrac{18000}{\sqrt{n}}\\\\n = (\dfrac{35280}{4200})^2 = (8.4)^2 = 70.56[/tex]

Since MOE is inversely proportional to root of n, thus, we will take sample as 71 which will make MOE to be less than (or say within, as asked in problem) $4200.

Thus,

The needed sample size for the given condition is 71

Learn more about margin or error here:

https://brainly.com/question/13990500