You are interested in the weight gain of the Aubum Freshman. You think that the

Question

You are interested in the weight gain of the Aubum Freshman. You think that the average weight gain of the Aubum student is not 15 pounds. You collect a random sample of measurements. The weight gains are (a negative value indicates a incempliehe anwer 16, 18, 16, 14, 6, 15, 14, 17, 15, 10, 8, 18, 21, 15, 14, 14,. 15, 9,5,8, 16, 17, 12, 15, 17 Yomay assure for the tok wing questions that the distribution of weight gain is normal (a) State the appropriate hypothesis test. (This was done before collecting the data) Null Hypothesis The weight gain is equal to 15 pounds F Fag question Altemative Hypothesis The weight gain is not equal to 15 pounds (b) Calculate the sample mean z. z= (c) Calculate the sample standard deviation s. s = (d) Evaluate the test statistic T= (e) The test statistic is t-distributed with ) The p-value for the test is (9) Should you reject the nul hypothesis in favor of the alternative at significance level a h) Is the weight gain significantly different from 15 ibs? No degrees of freedom 0.01? No

Explanation / Answer

Given that,
population mean(u)=15
on sample data
(
16, 18, 16, 14, 6, 15, 14, 17, 15, 10, 8, 18, 21, 15, 14, 14, 15, 9, 5, 8, 16, 17, 12, 15, 17
)
sample mean, x =13.8
standard deviation, s =4.0104
number (n)=25
null, Ho: =15
alternate, H1: !=15
level of significance, = 0.01
from standard normal table, two tailed t /2 =2.797
since our test is two-tailed
reject Ho, if to < -2.797 OR if to > 2.797
we use test statistic (t) = x-u/(s.d/sqrt(n))
to =13.8-15/(4.0104/sqrt(25))
to =-1.4961
| to | =1.4961
critical value
the value of |t | with n-1 = 24 d.f is 2.797
we got |to| =1.4961 & | t | =2.797
make decision
hence value of |to | < | t | and here we do not reject Ho
p-value :two tailed ( double the one tail ) - Ha : ( p != -1.4961 ) = 0.1477
hence value of p0.01 < 0.1477,here we do not reject Ho
ANSWERS
---------------
null, Ho: =15
alternate, H1: !=15
test statistic: -1.4961
critical value: -2.797 , 2.797
decision: do not reject Ho
p-value: 0.1477

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