Question # 10.2 Hints Sample size 20 n Population Mean 80 Sample Mean (X-bar) 85
ID: 3309493 • Letter: Q
Question
Question # 10.2 Hints Sample size 20 n Population Mean 80 Sample Mean (X-bar) 85 Sample Std. Dev. (S) 13.5 Std. Error (S/sqrt of n) Degrees of Freedom n-1 t-score (85-80)/std. error P-value (you can pick up a range of your p-value ) (?<your p-value<?) Check T-distribution in your textbook p. 530) with df=19; provide a interval estimate of your p-value (see a example in your textbook p. 181) How to interpret your p-value? Please refer to your textbook p. 181 (see the example to interpret the p-value) Step 1 Compute the Count Step2 Compute the Mean Step 3 Compute the Sample Std. Dev. Step 4 Compute the Std. Error - Sample Mean/SQRT of Count (N) Step 5 Determine the t-score based on the interval and levels of freedom Question # 10.2 Hints Sample size 20 n Population Mean 80 Sample Mean (X-bar) 85 Sample Std. Dev. (S) 13.5 Std. Error (S/sqrt of n) Degrees of Freedom n-1 t-score (85-80)/std. error P-value (you can pick up a range of your p-value ) (?<your p-value<?) Check T-distribution in your textbook p. 530) with df=19; provide a interval estimate of your p-value (see a example in your textbook p. 181) How to interpret your p-value? Please refer to your textbook p. 181 (see the example to interpret the p-value) Step 1 Compute the Count Step2 Compute the Mean Step 3 Compute the Sample Std. Dev. Step 4 Compute the Std. Error - Sample Mean/SQRT of Count (N) Step 5 Determine the t-score based on the interval and levels of freedomExplanation / Answer
10.2
a.
Given that,
population mean(u)=80
sample mean, x =85
standard deviation, s =13.5
number (n)=20
null, Ho: =80
alternate, H1: !=80
level of significance, = 0.05
from standard normal table, two tailed t /2 =2.093
since our test is two-tailed
reject Ho, if to < -2.093 OR if to > 2.093
we use test statistic (t) = x-u/(s.d/sqrt(n))
to =85-80/(13.5/sqrt(20))
to =1.656
| to | =1.656
critical value
the value of |t | with n-1 = 19 d.f is 2.093
we got |to| =1.656 & | t | =2.093
make decision
hence value of |to | < | t | and here we do not reject Ho
p-value :two tailed ( double the one tail ) - Ha : ( p != 1.6563 ) = 0.1141
hence value of p0.05 < 0.1141,here we do not reject Ho
ANSWERS
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null, Ho: =80
alternate, H1: !=80
test statistic: 1.656
critical value: -2.093 , 2.093
decision: do not reject Ho
p-value: 0.1141
b.
TRADITIONAL METHOD
given that,
sample mean, x =85
standard deviation, s =13.5
sample size, n =20
I.
stanadard error = sd/ sqrt(n)
where,
sd = standard deviation
n = sample size
standard error = ( 13.5/ sqrt ( 20) )
= 3.019
II.
margin of error = t /2 * (stanadard error)
where,
ta/2 = t-table value
level of significance, = 0.05
from standard normal table, two tailed value of |t /2| with n-1 = 19 d.f is 2.093
margin of error = 2.093 * 3.019
= 6.318
III.
CI = x ± margin of error
confidence interval = [ 85 ± 6.318 ]
= [ 78.682 , 91.318 ]
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DIRECT METHOD
given that,
sample mean, x =85
standard deviation, s =13.5
sample size, n =20
level of significance, = 0.05
from standard normal table, two tailed value of |t /2| with n-1 = 19 d.f is 2.093
we use CI = x ± t a/2 * (sd/ Sqrt(n))
where,
x = mean
sd = standard deviation
a = 1 - (confidence level/100)
ta/2 = t-table value
CI = confidence interval
confidence interval = [ 85 ± t a/2 ( 13.5/ Sqrt ( 20) ]
= [ 85-(2.093 * 3.019) , 85+(2.093 * 3.019) ]
= [ 78.682 , 91.318 ]
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interpretations:
1) we are 95% sure that the interval [ 78.682 , 91.318 ] contains the true population mean
2) If a large number of samples are collected, and a confidence interval is created
for each sample, 95% of these intervals will contains the true population mean