Question

You intend to estimate a population mean μμ with the following sample.

50.8

65

60

54.7

49.9

55.2

67.6

55.7

67.5

58.9

61.2

53.4

You believe the population is normally distributed. Find the 90% confidence interval. Enter your answer as an open-interval (i.e., parentheses) accurate to twp decimal places (because the sample data are reported accurate to one decimal place).

90% C.I. =

Answer 1

		x	(x-xbar)^2
		50.8	56.62562	Mean(x)=xbar=sum(x)/n	58.325
		65	2.805625	standard deviation(s)=sum(x-xbar)^2/n-1	6.088906
		60	13.14062	n	12
		54.7	70.98062	for 90 % confidence level with degree of freedom (n-1)=11
		49.9	9.765625	c	0.1
		55.2	86.02563	degrres of freedom	11
		67.6	6.890625	t=critical value obtain from t-table using df=n-1	1.795885
		55.7	84.18063	Margin of error =t*s/sqrt(n)	3.156655
		67.5	0.330625	LCL=xbar-ME	55.16835
		58.9	8.265625	UCL=xbar+ME	61.48165
		61.2	24.25562
		53.4	411618.5
	sum	699.9	411981.7

#the 90% confidence interval for μ

(xbar-ME,xbar+ME)

(55.1684,61.4817)

#the 90% confidence interval is  (55.1684,61.4817)