Consider a normal population with an unknown population standard deviation. A ra

Question

Consider a normal population with an unknown population standard deviation. A random sample results in :54.54 and s2-19.36. Use Table 2. a. Compute the 90% confidence interval for if and s2 were obtained from a sample of 12 observations. (Round intermediate calculations to 4 decimal places. Round "t" value to 3 decimal places and final answers to 2 decimal places.) Confidence interval to b. Compute the 90% confidence interval for if and s2 were obtained from a sample of 21 observations. (Round intermediate calculations to 4 decimal places. Round "t" value to 3 decimal places and final answers to 2 decimal places.) Confidence interval to c. Use your answers to discuss the impact of the sample size on the width of the interval. The bigger sample size will lead to a larger interval width and therefore a more precise interval. O The bigger sample size will lead to a smaller interval width and therefore a more precise interval

Explanation / Answer

Ans:

Given that

sample mean=54.54

sample variance,s2=19.36

a)when sample size,n=12

df=n-1=12-1=11

criticat t value=tinv(0.1,11)=1.796

90% confidence interval for population mean

=54.54+/-1.796*sqrt(19.36/12)

=54.54+/-2.28

=(52.26, 56.82)

b)when sample size,n=21

df=21-1=20

critical t value=tinv(0.1,20)=1.725

90% confidence interval for population mean

=54.54+/-1.725*sqrt(19.36/21)

=54.54+/-1.66

=(52.88, 56.20)

c)As,the sample size increased,margin of error decreases(from 2.28 to 1.66),so width of confidence interval decrease and hence more precise interval.

second option is correct.

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