Question

In this preliminary study at a level of significance of α-001 wa below 185 on the average? The deviation is, s 26.3. Assume t
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Answer #1

a)

One sided t test

H0; mu = 185
Ha: mu < 185

b)

test statistics;

t = (xbar -mu)/(s/sqrt(n)))
= ( 168.9 - 185)/(26.3/sqrt(6))
= -1.499


df = 6 -1 = 5

c)

p value = P(t < -1.499)

P value = 0.0970


Fail to reject H0 as p value > 0.01


d)

The t value at 99% confidence interval is,

alpha = 1 - 0.99 = 0.01
alpha/2 = 0.01/2 = 0.005
t(alpha/2,df) = t(0.005,5) = 4.032


Margin of error = E =z *(s/sqrt(n))
= 4.032 *(26.3/sqrt(6))
= 43.2913

CI = mean +/- E
= 168.9+/- 43.2913
= (125.61 , 212.19)

The 99% CI is (125.61 , 212.19)


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In this preliminary study at a level of significance of α-001 wa below 185 on the average? The deviation is, s 26.3. Assume the data is normally 5) after e mean cholesterol after the study is x=1...
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