Please show all work thank you Nutritional researchers condu

Please show all work, thank you.

Nutritional researchers conducted an investigation of two high-fiber diets intended to reduce serum cholesterol level. Twenty men with high serum cholesterol were randomly allocated to receive an oat diet or a bean diet for 21 days. The table summarizes the fall in serum cholesterol levels. (a) Is this a pooled or unpooled variance scenario? Justify your answer. (b) Based on your answer to part (a), construct a 90% confidence interval for the true average difference the fall in serum cholesterol levels between oat and bean diets among men with high serum cholesterol. Interpret this confidence interval. (c) Conduct a hypothesis test to assess whether or not there is a difference between oat and bean diets for fall in serum cholesterol levels. Use alpha = 0.05. For parts d and e, suppose that the researchers found an error found in the table above, and that the standard deviation n the bean diet sample was actually 12.5 instead of 29.4. (d) Based on this new information, construct a 99% confidence interval for the true average difference the fall in serum cholesteroll levels between oat and bean diets among men with high serum cholesterol. (e) Based on this new information, conduct a hypothesis test to assess whether or not there is a difference between oat and bean diets for fall n serum cholesterol levels. Use alpha = 0.05.

Solution

a)
Its an un-pooled variance scenerio.
b)
CI = x1 - x2 ± t a/2 * Sqrt ( sd1 ^2 / n1 + sd2 ^2 /n2 )
Where,
x1 = Mean of Sample 1, x2 = Mean of sample2
sd1 = SD of Sample 1, sd2 = SD of sample2
a = 1 - (Confidence Level/100)
ta/2 = t-table value
CI = Confidence Interval
Mean(x1)=53.6
Standard deviation( sd1 )=31.1
Sample Size(n1)=10
Mean(x2)=55.5
Standard deviation( sd2 )=29.4
Sample Size(n1)=10
CI = [ ( 53.6-55.5) ±t a/2 * Sqrt( 967.21/10+864.36/10)]
= [ (-1.9) ± t a/2 * Sqrt( 183.16) ]
= [ (-1.9) ± 3.25 * Sqrt( 183.16) ]
= [-45.88 , 42.08]

c)
Set Up Hypothesis
Null Hypothesis , There Is No-Significance between them Ho: u1 = u2
Alternate Hypothesis, There Is Significance between them - H1: u1 != u2
Test Statistic
X(Mean)=53.6
Standard Deviation(s.d1)=31.1 ; Number(n1)=10
Y(Mean)=55.5
Standard Deviation(s.d2)=29.4; Number(n2)=10
we use Test Statistic (t) = (X-Y)/Sqrt(s.d1^2/n1)+(s.d2^2/n2)
to =53.6-55.5/Sqrt((967.21/10)+(864.36/10))
to =-0.14
| to | =0.14
Critical Value
The Value of |t | with Min (n1-1, n2-1) i.e 9 d.f is 2.262
We got |to| = 0.14039 & | t | = 2.262
Make Decision
Hence Value of |to | < | t | and Here we Do not Reject Ho
P-Value: Two Tailed ( double the one tail ) - Ha : ( P != -0.1404 ) = 0.891
Hence Value of P0.05 < 0.891,Here We Do not Reject Ho