Question

Hey, could you please help me with this question. Please show me the steps. Thank you so much

For exercise 85, assume that a simple random sample has been selected and test the given claim. Use either the P-value method or the critical value method for testing hypothesis. Identify the nuil hypothesis, alternative hypothesis, test statistic, P-value (or range of P-values), critical value(s), and state the final conclusion that addresses the original claim. 8.5-Insomnia Treatment. A clinical trial was conducted to test the effectiveness of the drug Zopiclone for treating insomnia in older subjects. Before treatment with Zopiclone, 16 subjects had a mean wake time of 102.8 minutes. After treatment with Zopiclone, the 16 subjects had a mean wake time of 98.9 minutes and a standard deviation of 42.3 minutes (based on data from Cognitive Behavioral Therapy vs Zopiclone for Treatment of Chronic Primary Insomnia in Older Adults," by Siversten et al., Journal of the American Medical Association, Vol. 295, No. 24). Assume that the 16 sample values appear to be from a normally distributed population and test the claim that after treatment with Zopiclone, subjects have a mean wake tim of less than 102.8 minutes. Does Zopiclone appear to be effective? 1: Identify the null& alternative hypothesis. 2. To find the value of the test statistic of the given claim, follow the instructions below: a) Press STAT&scroll to TESTS & select "T-Test..." (option 2 in TI83Plus) & press Enter (for a hypothesis test of a claim involving one population). b) Highlight the option "Stats" and press ENTER. Then insert the following values: 1)Ho-claimed value of the population mean. This is the same value used in the null & alternative hypothesis. 2) x- sample mean.

Answer 1

8.5

Given that,
population mean(u)=102.8
sample mean, x =98.9
standard deviation, s =42.3
number (n)=16
null, Ho: μ=102.8
alternate, H1: μ<102.8
level of significance, alpha = 0.05
from standard normal table,left tailed t alpha/2 =1.753
since our test is left-tailed
reject Ho, if to < -1.753
we use test statistic (t) = x-u/(s.d/sqrt(n))
to =98.9-102.8/(42.3/sqrt(16))
to =-0.369
| to | =0.369
critical value
the value of |t alpha| with n-1 = 15 d.f is 1.753
we got |to| =0.369 & | t alpha | =1.753
make decision
hence value of |to | < | t alpha | and here we do not reject Ho
p-value :left tail - Ha : ( p < -0.3688 ) = 0.35872
hence value of p0.05 < 0.35872,here we do not reject Ho
ANSWERS
---------------
1.
null, Ho: μ=102.8
alternate, H1: μ<102.8
2.
test statistic: -0.369
3.
p-value: 0.35872
4.
critical value: -1.753
5.
decision: do not reject Ho
we fail to reject the null hypothesis that mean wake time is less than 102.8minutes