tes ) See what\'s Hot Web SIice Gallery Menstrual cycle lengths (days in an SRS

Question

tes ) See what's Hot Web SIice Gallery Menstrual cycle lengths (days in an SRS of nine women are as follows: (31, 28, 26, 24, 29, 33, 25, 26, 28). Use this data to test whether mean menstrual cycle length differs significantly from a lunar month (29.5 days). Assume the population values are Normal. Fill in each of the steps of the alpha level 0.05 two-sided hypothesis test: A. Hypothesis: H0 : -295 versus H1:A 29.5 B. Test Statistic: t obseved 1.78 Round to two decimal places C. P-value: Enter the range for the p-value found in Table C 0.20 D. Conclusion: Please enter Reject HO or "Fail to Reject HO his is case sensitive, Please type your answer rCanvas will not recognize it." Hint To complere this problem, you will need to compute the sample mean

Explanation / Answer

Given that,
population mean(u)=29.5
sample mean, x =27.778
standard deviation, s =2.9059
number (n)=9
null, Ho: =29.5
alternate, H1: !=29.5
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 =27.778-29.5/(2.9059/sqrt(9))
to =-1.78
| to | =1.78
critical value
the value of |t | with n-1 = 8 d.f is 2.306
we got |to| =1.78 & | 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 != -1.78 ) = 0.1133
hence value of p0.05 < 0.1133,here we do not reject Ho
ANSWERS
---------------
null, Ho: =29.5
alternate, H1: !=29.5
test statistic: -1.78
critical value: -2.306 , 2.306
decision: do not reject Ho, p-value: 0.1133 between 0.10 to 0.20

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