The average amount of rainfall during the summer months for the northeastern par
ID: 3319339 • Letter: T
Question
The average amount of rainfall during the summer months for the northeastern part of the United States is 11.52 inches. A researcher selects a random sample of 10 cities in the northeast and finds that the average amount of rainfall for 1995 was 7.42 inches. The sample standard deviation is 1.3 inches. At =0.05 can it be concluded that for 1995 the average rainfall was below 11.52 inches?
Summarize the data below:
=_____ n=_____
x¯=_____ s=_____
The hypotheses for this test are :
H0: =11.52 vs H1: ________ 11.52
Fill in the appropriate sign:
<
>
not equals
Enter the appropriate information into CrunchIt.
What P-value do you get?
1.000
<0.0001
Which of the following conclusions is best?
Fail to reject the null hypothesis.
Reject the null hypothesis
It is impossible to make a conclusion based on the P-Value.
4.The average amount of rainfall during the summer months for the northeastern part of the United States is 11.52 inches. A researcher selects a random sample of 10 cities in the northeast and finds that the average amount of rainfall for 1995 was 7.42 inches. The sample standard deviation is 1.3 inches. At =0.05 can it be concluded that for 1995 the average rainfall was below 11.52 inches?
Summarize the data below:
=_____ n=_____
x¯=_____ s=_____
The hypotheses for this test are :
H0: =11.52 vs H1: ________ 11.52
Fill in the appropriate sign:
<
>
not equals
Enter the appropriate information into CrunchIt.
What P-value do you get?
1.000
<0.0001
Which of the following conclusions is best?
Fail to reject the null hypothesis.
Reject the null hypothesis
It is impossible to make a conclusion based on the P-Value.
Explanation / Answer
4.
Given that,
population mean(u)=11.52
sample mean, x =7.42
standard deviation, s =1.3
number (n)=10
null, Ho: =11.52
alternate, H1: <11.52
level of significance, = 0.05
from standard normal table,left tailed t /2 =1.833
since our test is left-tailed
reject Ho, if to < -1.833
we use test statistic (t) = x-u/(s.d/sqrt(n))
to =7.42-11.52/(1.3/sqrt(10))
to =-9.973
| to | =9.973
critical value
the value of |t | with n-1 = 9 d.f is 1.833
we got |to| =9.973 & | t | =1.833
make decision
hence value of | to | > | t | and here we reject Ho
p-value :left tail - Ha : ( p < -9.9733 ) = 0
hence value of p0.05 > 0,here we reject Ho
ANSWERS
---------------
null, Ho: =11.52
alternate, H1: <11.52
test statistic: -9.973
critical value: -1.833
decision: reject Ho
p-value: 0 = <0.0001
we have enough evidence to support the claim that for 1995 the average rainfall was below 11.52 inches
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