Question

QUESTION PART A: You intend to estimate a population mean μμ from the following sample.

82.1	64	80	70.3
62.5	70.1	88.2	72.9
76.7	70.1	73.8	55.9
81	59.1	59.6	64
48.1	43.8	51	79
62.3	67.1	76	47.5
81.3	68.8	79.5	64.6
38.9	88.7	57.9	61.4
78.9	72.4	76.7	79.7
70	65.3	79.1	74.1

Find the 99.9% confidence interval. Enter your answer as a tri-linear inequality accurate to two decimal place (because the sample data are reported accurate to one decimal place).

_____ < μ < ____

QUESTION PART B: You intend to estimate a population mean μμ with the following sample.

35.2
34.1
9.9
40.2
35.7
36.7
54.9
30.7
38.1
39.6
23.1
65.9

You believe the population is normally distributed. Find the 98% 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).

98% C.I. =

Answer 1

Part A)

Sample size = n = 40

Sample mean = = 68.56

Standard deviation = s = 11.96904

We have to construct 99.9% confidence interval.

Formula is

Here E is a margin of error.

Degrees of freedom = n - 1 = 40 - 1 = 39

Level of significance = 0.001

tc = 3.558   ( Using t table)

So confidence interval is ( 68.56 - 6.7336 , 68.56 + 6.7336) = > ( 61.83 , 75.29)

61.83 < μ < 75.29

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Part B)

Sample size = n = 12

Sample mean = = 37.0083

Standard deviation = s = 13.9954

We have to construct 98% confidence interval.

Formula is

Here E is a margin of error.

Degrees of freedom = n - 1 = 12 - 1 = 11

Level of significance = 0.02

tc = 2.718 ( Using t table)

So confidence interval is ( 37.0083 - 10.9813 , 37.0083 + 10.9813) = > ( 26.03 , 47.99)