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Sagot :
Answer:
we conclude that mean DL reading for current smokers is significantly lower than 100 DL
μ ≤ 98.32
Step-by-step explanation:
Given :
H0 : μ = 100
H1 : μ < 100
Given the sample data:
103.768 88.602 73.003 123.086 91.052
92.295 61.675 90.677 84.023 76.014
100.615 88.017 71.210 82.115 89.222
102.754 108.579 73.154 106.755 90.479
Sample mean, xbar = 89.855
Sample standard deviation, s = 14.904
Test statistic :
(Xbar - μ) ÷ (s/√n)
(89.855 - 100) /(14.904/√20)
-10.145 / 3.3326357
Test statistic = - 3.044
Pvalue : using the Pvalue from T score calculator :
Pvalue = 0.00333
Pvalue < α ; Reject H0
Hence, we conclude that mean DL reading for current smokers is significantly lower than 100 DL.
b. Find a 99% upper one-sided confidence bound for the mean DL reading for current smokers. Does this bound confirm your conclusions in part a?
Tcritical at 99% (one sided) = 2.54
Upper confidence interval :
Xbar ± standard Error
Standard Error = Tcritical * s/√n
Standard Error = 2.54 * 14.904/√20 = 8.465
Upper boundary:
Xbar + standard error
89.855 + 8.465 = 98.32
μ ≤ 98.32
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