Question

A simple random sample of size n is drawn from a population that is normally distributed. The sample​ mean,

x overbarx​,

is found to be

108108​,

and the sample standard​ deviation, s, is found to be

1010.

​(a) Construct

aa

9898​%

confidence interval about

muμ

if the sample​ size, n, is

2121.

​(b) Construct

aa

9898​%

confidence interval about

muμ

if the sample​ size, n, is

1414.

​(c) Construct

aa

9696​%

confidence interval about

muμ

if the sample​ size, n, is

2121.

​(d) Could we have computed the confidence intervals in parts​ (a)-(c) if the population had not been normally​ distributed?

Answer 1

Answer:

Given that ,

mean x = 108 , standard deviation s = 10

(a) here given that n = 21 then df = n - 1 = 21 - 1 = 20

the value of 98% confidence with df from t table is t = 2.528

confidence interval formula

=> mean +/- t * s/sqrt(n)

=> 108 +/- 2.528 * 10/sqrt(21)

= 108 +/- 5.516

=> (102.484 , 113.516)

lower bound = 102.484

upper bound = 113.516

(b) here given that n = 14 then df = n - 1 = 14 - 1 = 13

the value of 90% confidence with df from t table is t = 2.650

confidence interval formula

=> mean +/- t * s/sqrt(n)

=> 108 +/- 2.650 * 10/sqrt(14)

= 108 +/- 7.082

=> (100.918 , 115.082)

lower bound = 100.918

upper bound = 115.082

(c)here given that n = 21 then df = n - 1 = 21 - 1 = 20

the value of 70% confidence with df from t table is t = 2.086

confidence interval formula

=> mean +/- t * s/sqrt(n)

=> 108 +/- 2.086 * 10/sqrt(241)

= 108+/- 4.552

=> (103.448, 112.552)

lower bound = 103.448

upper bound = 112.552

(d)if the population had not been normally​ distributed

=> No, the population needs to be normally distributed

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