The mean starting salaries of a random sample of three students who saduated fro

Question

The mean starting salaries of a random sample of three students who saduated from college last year with majors in the mathematical sciences was $44,000 with standard error $5773.51 (Note: do not use $ symbols or commas in your answers.) (a) what is the margin of error for a 95% confidence interval? (Round to the nearest whole number.) and the upper limit of the 95% confidence interval. (Round to the (b) Find the lower limit nearest whole number.) (c) Other than decreasing the level of confd O decrease the sample size O increase the sample size O decrease the population mean O decrease the sample mean (d) Which of the two statements below must be reasonably true for the interval to be valid? The sample size is large. There are at least 15 successes and 15 failures The sample was randomly selected. The population is approximately normal. Submit Answer Save Progress

Explanation / Answer

The formula for estimation is:

= M ± Z(sM)

where:

M = sample mean
Z = Z statistic determined by confidence level
sM = standard error = (s2/n)

Calculation -

M = 44000
t = 1.96
sM = (5773.512/3) = 3333.34

= M ± Z(sM)
= 44000 ± 1.96*3333.34
= 44000 ± 6533.22

Result -

M = 44000, 95% CI [37466.78, 50533.22].

You can be 95% confident that the population mean () falls between 37466.78 and 50533.22.

Note - [ Margin of Error: 6500 (if a rounded answer is needed or else 6533.22 is the accurate answer.]

(a) Margin of error - 6533

(b) Lower limit - 37467, Upper limit - 50533

(c) Increase the sample size.

(d) The sample was randomly selected.

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