(Round all intermediate calculations to at least 4 decimal places.) Consider the

Question

(Round all intermediate calculations to at least 4 decimal places.)

Consider the following hypotheses:

H0:  = 8HA:   8

The population is normally distributed. A sample produces the following observations:

       6 9 8 7 7 1 1 1 0

Use the p-value approach to conduct the test at a 5% level of significance. Use Table 2.

 Click here for the Excel Data File

a.

Find the mean and the standard deviation. (Round your answers to 2 decimal places.)

    Mean    Standard deviation  
    8.29       1.80

b.

Calculate the value of the test statistic. (Round your answer to 2 decimal places.)

  Test statistic  

c.

Approximate the p-value of the test statistic.

   p-value < 0.020.02 < p-value < 0.050.05 < p-value < 0.100.10 < p-value < 0.20p-value > 0.20

d.What is the conclusion?   Reject H0 since the p-value is greater than .Reject H0 since the p-value is smaller than .Do not reject H0 since the p-value is greater than .Do not reject H0 since the p-value is

Ive answered most of the question (A , C, D ) except for the test statistic which is part B.    I thought it was .44 or 1.44 but those were both incorrect.

Explanation / Answer

a.
Given that,
population mean(u)=8
sample mean, x =4.44
standard deviation, s =3.6094
number (n)=9
null, Ho: =8
alternate, H1: !=8
level of significance, = 0.05
from standard normal table, two tailed t /2 =2.306
since our test is two-tailed
reject Ho, if to < -2.306 OR if to > 2.306
we use test statistic (t) = x-u/(s.d/sqrt(n))
to =4.44-8/(3.6094/sqrt(9))
to =-2.9589
| to | =2.9589
critical value
the value of |t | with n-1 = 8 d.f is 2.306
we got |to| =2.9589 & | t | =2.306
make decision
hence value of | to | > | t | and here we reject Ho
p-value :two tailed ( double the one tail ) - Ha : ( p != -2.9589 ) = 0.0182
hence value of p0.05 > 0.0182,here we reject Ho
ANSWERS
---------------
null, Ho: =8
alternate, H1: !=8
test statistic: -2.9589
critical value: -2.306 , 2.306
decision: reject Ho
p-value: 0.0182
b.
Given that,
population mean(u)=8
sample mean, x =8.29
standard deviation, s =1.8
number (n)=9
null, Ho: =8
alternate, H1: !=8
level of significance, = 0.05
from standard normal table, two tailed t /2 =2.306
since our test is two-tailed
reject Ho, if to < -2.306 OR if to > 2.306
we use test statistic (t) = x-u/(s.d/sqrt(n))
to =8.29-8/(1.8/sqrt(9))
to =0.48
| to | =0.48
critical value
the value of |t | with n-1 = 8 d.f is 2.306
we got |to| =0.48 & | t | =2.306
make decision
hence value of |to | < | t | and here we do not reject Ho
p-value :two tailed ( double the one tail ) - Ha : ( p != 0.4833 ) = 0.6418
hence value of p0.05 < 0.6418,here we do not reject Ho
ANSWERS
---------------
null, Ho: =8
alternate, H1: !=8
test statistic: 0.48
critical value: -2.306 , 2.306
decision: do not reject Ho
c.
p-value: 0.6418

d.
Do not reject H0 since the p-value is greater than

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