Question

A.)

A medical researcher is studying the effects of a drug on blood pressure. Subjects in the study have their blood pressure taken at the beginning of the study. After being on the medication for 4 weeks, their blood pressure is taken again. The change in blood pressure is recorded and used in doing the hypothesis test.

Change: Final Blood Pressure - Initial Blood Pressure

The researcher wants to know if there is evidence that the drug increases blood pressure. At the end of 4 weeks, 36 subjects in the study had an average change in blood pressure of 2.4 with a standard deviation of 4.5.

Find the p-value for the hypothesis test. ___________

Your answer should be rounded to 4 decimal places.

B.)

Find the p-value for the hypothesis test. A random sample of size 54 is taken. The sample has a mean of 426 and a standard deviation of 82.

H0: µ = 400

Ha: µ ≠ 400

The p-value for the hypothesis test is . ________________

Your answer should be rounded to 4 decimal places.

C.)

Child Health and Development Studies (CHDS) has been collecting data about expectant mothers in Oakland, CA since 1959. One of the measurements taken by CHDS is the weight increase (in pounds) for expectant mothers in the second trimester.

In a fictitious study, suppose that CHDS finds the average weight increase in the second trimester is 14 pounds. Suppose also that, in 2015, a random sample of 43 expectant mothers have mean weight increase of 16.2 pounds in the second trimester, with a standard deviation of 5.7 pounds.

A hypothesis test is done to see if there is evidence that weight increase in the second trimester is greater than 14 pounds.

Find the p-value for the hypothesis test. _____________________

The p-value should be rounded to 4 decimal places.

I need to know how to do these with excel but regular way works too if its too much.

Answer 1

A.
Assume values
population mean(u)=8.4
sample mean, x =6
here Assume values are taking because given problem only change blood pressure (final -initial).
standard deviation, s =4.5
number (n)=36
null, Ho: μ=8.4
alternate, H1: μ!=8.4
level of significance, alpha = 0.05
from standard normal table, two tailed t alpha/2 =2.03
since our test is two-tailed
reject Ho, if to < -2.03 OR if to > 2.03
we use test statistic (t) = x-u/(s.d/sqrt(n))
to =6-8.4/(4.5/sqrt(36))
to =-3.2
| to | =3.2
critical value
the value of |t alpha| with n-1 = 35 d.f is 2.03
we got |to| =3.2 & | t alpha | =2.03
make decision
hence value of | to | > | t alpha| and here we reject Ho
p-value :two tailed ( double the one tail ) - Ha : ( p != -3.2 ) = 0.0029
hence value of p0.05 > 0.0029,here we reject Ho
ANSWERS
---------------
null, Ho: μ=8.4
alternate, H1: μ!=8.4
test statistic: -3.2
critical value: -2.03 , 2.03
decision: reject Ho
p-value: 0.0029
we have enough evidence to support the claim that the drug increases blood pressure

B.
Given that,
population mean(u)=400
sample mean, x =426
standard deviation, s =82
number (n)=54
null, Ho: μ=400
alternate, H1: μ!=400
level of significance, alpha = 0.05
from standard normal table, two tailed t alpha/2 =2.006
since our test is two-tailed
reject Ho, if to < -2.006 OR if to > 2.006
we use test statistic (t) = x-u/(s.d/sqrt(n))
to =426-400/(82/sqrt(54))
to =2.33
| to | =2.33
critical value
the value of |t alpha| with n-1 = 53 d.f is 2.006
we got |to| =2.33 & | t alpha | =2.006
make decision
hence value of | to | > | t alpha| and here we reject Ho
p-value :two tailed ( double the one tail ) - Ha : ( p != 2.33 ) = 0.0236
hence value of p0.05 > 0.0236,here we reject Ho
ANSWERS
---------------
null, Ho: μ=400
alternate, H1: μ!=400
test statistic: 2.33
critical value: -2.006 , 2.006
decision: reject Ho
p-value: 0.0236
we have enough evidence to support the claim

C.
Given that,
population mean(u)=14
sample mean, x =16.2
standard deviation, s =5.7
number (n)=43
null, Ho: μ=14
alternate, H1: μ>14
level of significance, alpha = 0.05
from standard normal table,right tailed t alpha/2 =1.682
since our test is right-tailed
reject Ho, if to > 1.682
we use test statistic (t) = x-u/(s.d/sqrt(n))
to =16.2-14/(5.7/sqrt(43))
to =2.5309
| to | =2.5309
critical value
the value of |t alpha| with n-1 = 42 d.f is 1.682
we got |to| =2.5309 & | t alpha | =1.682
make decision
hence value of | to | > | t alpha| and here we reject Ho
p-value :right tail - Ha : ( p > 2.5309 ) = 0.0076
hence value of p0.05 > 0.0076,here we reject Ho
ANSWERS
---------------
null, Ho: μ=14
alternate, H1: μ>14
test statistic: 2.5309
critical value: 1.682
decision: reject Ho
p-value: 0.0076
we have enough evidence to support the claim that weight increase in the second trimester is greater than 14 pounds