9.2.17 A simple random sample of size n is drawn trom a population that is norma

Question

9.2.17 A simple random sample of size n is drawn trom a population that is normally distributed The sample mean, k is found to be 107, and the (b) Construct an 80% (d) Codd we have computed te cont dence ntervas in parts (?) (c) if the populaton had not been normally distributed? sample standard devation, s, s found lo be t0 interval about u id the sample size, n, is 29 Click the icon to view the table of areas under the t-distrbution (a) C an80% confidence nterval about ? fthe sample size, n, is 20 Lower bound Depper bound ? (Use ascending order Round to one decimal place as needed ) 5 O Type here to search

Explanation / Answer

A)

mean= 107

sd= 10

n= 20

alpha= 0.2

t(a/2,n-1) t(0.2/2,20-1) 1.328

CI = mean +- t(a/2,n-1)*(sd/sqrt(n))

lower = 107 - 1.3277282090268*(10/sqrt(20))= 104.0

upper = 107 + 1.3277282090268*(10/sqrt(20))= 110.0

B)

mean= 107

sd= 10

n= 29

alpha= 0.2

t(a/2,n-1) t(0.2/2,29-1) 1.313

CI = mean +- t(a/2,n-1)*(sd/sqrt(n))

lower = 107 - 1.31252678159267*(10/sqrt(29))= 104.6

upper = 107 + 1.31252678159267*(10/sqrt(29))= 109.4

C)

mean= 107

sd= 10

n= 20

alpha= 0.02

t(a/2,n-1)= t(0.02/2,20-1) =2.539

CI = mean +- t(a/2,n-1)*(sd/sqrt(n))

lower = 107 - 2.53948319062396*(10/sqrt(20))= 101.3

upper = 107 + 2.53948319062396*(10/sqrt(20))= 112.7

D)

The above confidence intervals would not have been completed if the population had not be normally distributed. the assumption for applicability of one sample T test is that the population from which samples are drawn is normally distributed

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