# For a random sample of 18 rece

For a random sample of 18 recent business school graduatesbeginning their first job, the mean starting salary was found to be\$39,500, and the sample standard deviation was \$9,500. Assuming thepopulation is normally distributed, find the margin of error of a95% confidence interval for the population mean.

Answer:

Solution :

Given that,

Point estimate = sample mean = = 39500

sample standard deviation = s = 9500

sample size = n = 18

Degrees of freedom = df = n – 1 = 18 – 1 = 17

At 95% confidence level the t is ,

= 1 – 95% = 1 – 0.95 = 0.05

/ 2 = 0.05 / 2 = 0.025

t/2,df = t0.025,17 = 2.110

Margin of error = E = t/2,df* (s /$\sqrt$n)

= 2.110 * (9500 / $\sqrt$18)

= 4724.652

Margin of error is 4724.652

The 95% confidence interval estimate of the population meanis,

– E < < + E

39500 – 4724.652 < < 39500 + 4724.652

34775.348 < < 44224.652

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