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MATH 225N Week 7 Assisgnment / MATH225 Week7 Assisgnment : Conducting a hypothesis test P-Value Approach(Version 5, New-2021) :Chamberlain College of Nursing (ANSWERS VERIFIED) $10.49
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MATH 225N Week 7 Assisgnment / MATH225 Week7 Assisgnment : Conducting a hypothesis test P-Value Approach(Version 5, New-2021) :Chamberlain College of Nursing (ANSWERS VERIFIED)
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MATH 225N / MATH225
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Chamberlain College Of Nursing University
MATH 225N Week 7 Assisgnment / MATH225 Week7 Assisgnment : Conducting a hypothesis test P-Value Approach(Version 5, New-2021) :Chamberlain College of Nursing (ANSWERS VERIFIED)
math 225n week 7 assisgnment math225 week7 assisgnment conducting a hypothesis test p value approachversion 5
new 2021 chamberlain college of nursing answers verified
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MATH225 Week7 Assignment
(Conducting a hypothesis test P-Value Approach)
Determine the p-value for a hypothesis test for the mean (population standard deviation known)
Question
What is the p-value of a two-tailed one-mean hypothesis test, with a test statistic
of z0=−1.73? (Do not round your answer; compute your answer using a value
from the table below.) z−1.8−1.7−1.6−1.5−1.4 .......
Great work! That's correct.
0.084
Answer Explanation
Correct answers:
0.084
The p-value is the probability of an observed value of z=1.73 or greater in
magnitude if the null hypothesis is true, because this hypothesis test is two-tailed.
This means that the p-value could be less than z=−1.73, or greater
than z=1.73. This probability is equal to the area under the Standard Normal
curve that lies either to the left of z=−1.73, or to the right of z=1.73.
A normal curve is over a horizontal axis and is centered on 0. Two points are
labeled negative 1.73 and 1.72 The area to the right of 1.73 and to the left of
negative 1.73 is shaded.
Using the Standard Normal Table given, we can see that the p-value that
corresponds with z=−1.73 is 0.042, which is just the area to the left
of z=−1.73. Since the Standard Normal curve is symmetric, the area to the right
, of z=1.73 is 0.042 as well. So, the p-value of this two-tailed one-mean
hypothesis test is (2)(0.042)=0.084
Make a conclusion and interpret the results of a one-mean hypothesis test (population standard
deviation known) using the P-Value Approach
Question
Mary, a javelin thrower, claims that her average throw is 61 meters. During a practice session,
Mary has a sample throw mean of 55.5 meters based on 12 throws. At the 1% significance
level, does the data provide sufficient evidence to conclude that Mary's mean throw is less
than 61 meters? Accept or reject the hypothesis given the sample data below.
Reject the null hypothesis because |−1.99|>0.01.
Do not reject the null hypothesis because |−1.99|>0.01.
Reject the null hypothesis because the p-value 0.0233 is greater than the
significance level α=0.01.
Do not reject the null hypothesis because the value of z is negative.
Do not reject the null hypothesis because the p-value 0.0233 is greater than
the significance level α=0.01.
Answer Explanation
Correct answer:
Do not reject the null hypothesis because the p-value 0.0233 is greater than the
significance level α=0.01.
In making the decision to reject or not reject H0, if α>p-value, reject H0 because the
results of the sample data are significant. There is sufficient evidence to conclude
that H0 is an incorrect belief and that the alternative hypothesis, Ha, may be correct.
If α≤p-value, do not reject H0. The results of the sample data are not significant, so
there is not sufficient evidence to conclude that the alternative hypothesis, Ha, may be
correct. In this case, α=0.01 is less than or equal to p=0.0233, so the decision
is to not reject the null hypothesis.
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