Consider the following hypothesis test: H 0 : = 18 H a : 18 A sample of 48 provi

Question

Consider the following hypothesis test:

H0: = 18
Ha: 18

A sample of 48 provided a sample mean = 17 and a sample standard deviation s = 4.5.

If requires, round your answers to two decimal places.

a. Compute the value of the test statistic.

b. Use the t distribution table (Table 2 in Appendix B) to compute a range for the p-value.

p-value is between and

c. At = .05, what is your conclusion?

p-value Selectgreater than or equal to 0.05, rejectgreater than 0.05, do not rejectless than or equal to 0.05, do not rejectless than 0.05, rejectequal to 0.05, do not rejectnot equal to 0.05, do not rejectItem 4 H0

d. What is the rejection rule using the critical value?

Reject H0 if t Selectgreater than or equal to -2.012greater than 2.012less than or equal to -2.012less than -2.012equal to 2.012not equal to -2.012Item 5 or t Selectgreater than or equal to 2.012greater than -2.012less than or equal to 2.012less than -2.012equal to 2.012not equal to -2.012Item 6

What is your conclusion?

t = ; Selectdo not rejectrejectItem 8 H0

Explanation / Answer

Given that,
population mean(u)=18
sample mean, x =17
standard deviation, s =4.5
number (n)=48
null, Ho: =18
alternate, H1: !=18
level of significance, = 0.05
from standard normal table, two tailed t /2 =2.012
since our test is two-tailed
reject Ho, if to < -2.012 OR if to > 2.012
we use test statistic (t) = x-u/(s.d/sqrt(n))
to =17-18/(4.5/sqrt(48))
to =-1.54
| to | =1.54
critical value
the value of |t | with n-1 = 47 d.f is 2.012
we got |to| =1.54 & | t | =2.012
make decision
hence value of |to | < | t | and here we do not reject Ho
p-value :two tailed ( double the one tail ) - Ha : ( p != -1.5396 ) = 0.1304
hence value of p0.05 < 0.1304,here we do not reject Ho
ANSWERS
---------------
null, Ho: =18
alternate, H1: !=18
test statistic: -1.54
critical value: -2.012 , 2.012
decision: do not reject Ho
p-value: 0.1304

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