Question

In: Statistics and Probability

Construct a 98​% confidence interval to estimate the population mean when x=65 and s​=14.7 for the...

Construct a 98​% confidence interval to estimate the population mean when x=65 and s​=14.7 for the sample sizes below. ​

a) n=19 ​

b) n=43

​c) n=58 ​

a) The 98​% confidence interval for the population mean when n=19 is from a lower limit of to an upper limit of . ​(Round to two decimal places as​ needed.)

​b) The 98​% confidence interval for the population mean when n=43 is from a lower limit of to an upper limit of. ​(Round to two decimal places as​ needed.)

​c) The 98​% confidence interval for the population mean when n=58 is from a lower limit of to an upper limit of . ​(Round to two decimal places as​ needed.)

Solutions

Expert Solution

Solution :

Given that,

Point estimate = sample mean = = 65

sample standard deviation = s = 14.7

a)

sample size = n = 19

Degrees of freedom = df = n - 1 = 18

At 98% confidence level the t is ,

= 1 - 98% = 1 - 0.98 = 0.02

/ 2 = 0.02 / 2 = 0.01

t /2,df = t0.01,18 = 2.552

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

= 2.552* (14.7 / 19)

= 8.61

The 98% confidence interval estimate of the population mean is,

- E < < + E

65 - 8.61 < < 65 + 8.61

56.39 < < 73.61

The 98​% confidence interval for the population mean when n=19 is from a lower limit of 56.39 to an upper limit of 73.61.

b)

sample size = n = 43

Degrees of freedom = df = n - 1 = 42

At 98% confidence level the t is ,

= 1 - 98% = 1 - 0.98 = 0.02

/ 2 = 0.02 / 2 = 0.01

t /2,df = t0.01,42 = 2.416

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

= 2.416* (14.7 / 43)

= 5.42

The 98% confidence interval estimate of the population mean is,

- E < < + E

65 - 5.41 < < 65 + 5.41

59.59 < < 70.41

The 98​% confidence interval for the population mean when n=43 is from a lower limit of 59.59 to an upper limit of 70.41.

c)

sample size = n = 58

Degrees of freedom = df = n - 1 = 57

At 98% confidence level the t is ,

= 1 - 98% = 1 - 0.98 = 0.02

/ 2 = 0.02 / 2 = 0.01

t /2,df = t0.01,57 = 2.394

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

= 2.394* (14.7 / 58)

= 4.62

The 98% confidence interval estimate of the population mean is,

- E < < + E

65 - 4.62 < < 65 + 4.62

60.38 < < 69.62

The 98​% confidence interval for the population mean when n=58 is from a lower limit of 60.38 to an upper limit of 69.62.


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