The following are the scores of 15 randomly selected students in
a national test. Assume the distribution of scores is normal. Use
the sample data to construct a 95% confidence interval for the mean
of scores for the population.
8.6; 9.4; 7.9; 6.8; 8.3; 7.3; 9.2; 9.6; 8.7; 11.4; 10.3; 5.4; 8.1;
5.5; 6.9
Select one:
A. 7.3006< p<9.1528
B. 7.3006< μ<9.1528
C. 7.3804< p<9.0729
D. 7.3804< μ<9.0729
sample mean, xbar = 8.2267
sample standard deviation, s = 1.6722
sample size, n = 15
degrees of freedom, df = n - 1 = 14
Given CI level is 95%, hence α = 1 - 0.95 = 0.05
α/2 = 0.05/2 = 0.025, tc = t(α/2, df) = 2.145
ME = tc * s/sqrt(n)
ME = 2.145 * 1.6722/sqrt(15)
ME = 0.926
CI = (xbar - tc * s/sqrt(n) , xbar + tc * s/sqrt(n))
CI = (8.2267 - 2.145 * 1.6722/sqrt(15) , 8.2267 + 2.145 *
1.6722/sqrt(15))
CI = (7.3006 , 9.1528)
B. 7.3006< μ<9.1528
The following are the scores of 15 randomly selected students in a national test. Assume the...
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