Question

Consider a normal population with an unknown population standard deviation. A random sample results in x = 40.62 and s2 - 21.16. [You may find it useful to reference the t table.] a. Compute the 99% confidence interval for u if x and s2 were obtained from a sample of 8 observations. (Round intermediate calculations to at least 4 decimal places. Round "to" value to 3 decimal places and final answers to 2 decimal places.) Confidence intervalſ to b. Compute the 99% confidence interval for u if x and s2 were obtained from a sample of 13 observations. (Round intermediate calculations to at least 4 decimal places. Round "t" value to 3 decimal places and final answers to 2 decimal places.) Confidence interval

Answer 1

a) Mean ($\bar{x}$) = 40.62
Sample size (n) = 8
Standard deviation (s) = 4.6
Confidence interval(in %) = 99
t = 3.499

ta/2.n-1 = 3.499

Since we know that
Confidence\; interval =
u l-u'7/07 72 S

Required confidence interval =
(40.62 -3.499: = 40.62 +3.4991:

Required confidence interval = (40.62-5.6906, 40.62+5.6906)
Required confidence interval = (34.9294, 46.3106)

b) Mean ($\bar{x}$) = 40.62
Sample size (n) = 13
Standard deviation (s) = 4.6
Confidence interval(in %) = 99
t = 3.0545

ta/2,1-1 = 3.0545

Required confidence interval =
(40.62 -3.0545 40.62 + 3.05.45 4 ) 13

Required confidence interval = (40.62-3.897, 40.62+3.897)
Required confidence interval = (36.723, 44.517)
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