9.2.17 Question Help A simple random sample of size n is drawn from a population

Question

9.2.17 Question Help A simple random sample of size n is drawn from a population that is normally distributed. The sample mean, x, is found to be 110, and the sample standard deviation, s, is found to be 10 (a) Construct a 90% confidence interval about if the sample size, n, is 18. (b) Construct a 90%confidence interval about if the sample size, n, is 24 (c) Construct a 98% confidence interval about if the sample size, n, is 18. (d) Could we have computed the confidence intervals in parts (a)Hc) if the population had not been normally distributed? Click the icon to view the table of areas under the t-distribution.

Explanation / Answer

Here the sample mean x = 110

sample standard deviation s = 10

so,

standard error of sample mean = s/n = 10/ n

(a) Here if n = 18

90% confidence interval = x +- tdf, )/2   s/n

dF = 18 -1 = 17, = 1 -0.90 = 0.1

so t17, 0.05 = 1.7396

so putting values

90% confidence interval = x +- tdf, )/2   s/n = 110 + 1.7396 * 10 /18

90% confidence interval = 110 +- 4.100 = (105. 900 , 114.100)

(b) Here if n = 24

90% confidence interval = x +- tdf, /2   s/n

dF = 24 -1 = 23, = 1 -0.90 = 0.1

so t23, 0.05 = 1.7171

so putting values

90% confidence interval = x +- tdf, )/2   s/n = 110 + 1.7139 * 10 /24

90% confidence interval = 110 +- 3.498 = (106.502 , 113.498)

(c) Here if n = 18

90% confidence interval = x +- tdf, /2   s/n

dF = 18 -1 = 17, = 1 -0.98 = 0.02

so t17, 0.01 = 2.5669

so putting values

98% confidence interval = x +- tdf, )/2   s/n = 110 + 2.5669 * 10 /18

98% confidence interval = 110 +- 6.050 = (103.950 , 116.050)

(d) Yes, if we don't have data that are not normally distributed. we can calculate it with the help of t- distriubtion as given above in similar method.

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