A researcher is interested in studying the effects of a new

A researcher is interested in studying the effects of a new treatment for bipolar disorder. In the population, bipolar disorder has a mean of 30 (based on a scale ranging from 0-100; higher numbers indicate a greater number of symptoms). She randomly selects 60 patients to participate in her study. After two weeks she obtains the following data:

M = 26.4

s2 = 145.8

a. What type of test will she need to conduct?

b. Test the effectiveness of her new treatment using the steps of Null Hypothesis testing

c. Calculate the 95% confidence interval for these data. Make sure you interpret as well.

Please use what\'s below, please.

Solution

a)
t-test
b)
Set Up Hypothesis
Null Hypothesis H0: U=30
Alternate Hypothesis H1: U!=30
Test Statistic
Population Mean(U)=30
Sample X(Mean)=26.4
Standard Deviation(S.D)=12.075
Number (n)=60
we use Test Statistic (t) = x-U/(s.d/Sqrt(n))
to =26.4-30/(12.075/Sqrt(59))
to =-2.309
| to | =2.309
Critical Value
The Value of |t | with n-1 = 59 d.f is 2.001
We got |to| =2.309 & | t | =2.001
Make Decision
Hence Value of | to | > | t | and Here we Reject Ho
P-Value :Two Tailed ( double the one tail ) -Ha : ( P != -2.3094 ) = 0.0244
Hence Value of P0.05 > 0.0244,Here we Reject Ho

c)
Confidence Interval
CI = x ± t a/2 * (sd/ Sqrt(n))
Where,
x = Mean
sd = Standard Deviation
a = 1 - (Confidence Level/100)
ta/2 = t-table value
CI = Confidence Interval
Mean(x)=26.4
Standard deviation( sd )=12.075
Sample Size(n)=60
Confidence Interval = [ 26.4 ± t a/2 ( 12.075/ Sqrt ( 60) ) ]
= [ 26.4 - 2.001 * (1.56) , 26.4 + 2.001 * (1.56) ]
= [ 23.28,29.52 ]