5. Suppose the smple mean for a normally distributed population is 198 and the s

Question

5. Suppose the smple mean for a normally distributed population is 198 and the sample size is 40. Find a 95% confidence interviu for the population mean, for each of the blowing situations (a) The true population standard deviation is a 20.0. standard error crit, value Co/2 or Lo)mrgn of Error ,1981 1,10, 312.3 :le,1931 2 0 40 95% Confidence interval dil 8 , 204.2 ) (b) The true population standard deviation is not known and the sample standard deviation, is found to be s = 18.3 standard error crit. value (Co/2 or (o/2) margin of Error E 2,83 2,623S8S3S p(-2.02 3L+3112,023) 95% confidence interval (112·203.9 )

Explanation / Answer

given that sample size = 40

so from the z table we see that the z score for 99% CI is 2.58

SD = 20

so SE = SD/sqrt(n)
= 20/sqrt(40) = 3.162

also the z critcal value is 2.58

so the Margin of error is
MOE = Z*SE
= 2.58*3.162
= 8.157

now the Xbar-E = 198-8.157 = 189.84

Xbar+E = 198+8.157 = 206.15

b)

given that sample size = 40

so from the z table we see that the z score for 99% CI is 2.58

SD = 20

so SE = s/sqrt(n)
= 18.3/sqrt(40) = 2.893

also the z critcal value is 2.58

so the Margin of error is
MOE = Z*SE
= 2.58*2.893
= 7.463

now the Xbar-E = 198-7.463 = 190.53

Xbar+E = 198+7.463 = 205.46

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