Please use the following problems as your Module 6 Discussion Questions:
page 372, ##20,24
page 383, #10
I DONT HAVE BOOK SO PLEASE COMPILE AN ANSWER BASED ON MY PEER’S ANSWERS BELOW:
372 # 20, 24
20. P values.
A. ID the hypothesis test as being two tailed, left or right tailed
B. Find the P value
C. Using a significance level of a=.05 should we reject or fail to reject null
20. The test statistic of Z= -1.94 is obtained when testing the claim that p=3/8
H0: P=3/8 or .375
H1: P≠3/8 or .375
H1< left tail) H1> right tail) H1≠ two tailed
Normalcdf ( -99999, -1.94 = .0262
Two tails .0262*2
P-Value=.0524
.0524 > .05 SO fail to reject
24. Critical values
A. Find the critical values
B. LOS a=.05 should we reject or fail to reject null
24. Exercise 20
Critical region -1.94 from 20
T test
Divide .05 by 2=.025
Invonorm (.025= -1.96
1-.025=1.96
Fail to reject because -1.96 is not inside the critical region -1.94
383 #10 test the given claim. A. Identify the null hypothesis, alternative hypothesis, B. test statistic, P-value, or critical value(s), then C. state the conclusion about the null hypothesis, as well as the final conclusion that addresses the original claim. Use the P-value method unless your instructor specifies otherwise. Use the normal distribution as an approximation to the binomial distribution, as described in Part 1 of this section.
Eliquis: The drug Eliquis (apixaban) is used to help prevent blood clots in certain patients. In clinical trials, among 5924 patients treated with Eliquis, 153 developed the adverse reaction of nausea (based on data from Bristol Myers Squibb Co.). Use a .05 significance level to test the claim that 3% of Eliquis users develop nausea. Does nausea appear to be a problematic adverse reaction?
Claim: U=.03 n=5924 xbar= 153 a=.05
H0 p=.03 1 prop Ztest
H1 P≠.03 P=.0597> .05 So fail to reject
Conclusion is that the mean 3% of Eliquis users develop nausea