a. Standard error 1. For each group, what is the standard error of the mean fall

Question

a. Standard error

1. For each group, what is the standard error of the mean fall in cholesterol?

2. Imagine that we studied 50 men in each group instead of 10. Would you expect the SD for each group to change, and if so, how? Would you expect the SE for each group mean to change, and if so, how? How does this illustrate the difference between SD and SE?

3. What is the standard error of the difference in means between the two groups?

B. Let’s find out whether the oat diet reduces cholesterol in this population on average.

1. What kind of sample do we have? Two independent, single or paired?

2. What kind of question are we asking? Two-sided/nondirectional or one- sided/directional?

3. State the null and alternative hypotheses.

4. Write out and calculate the t-statistic.

5. What is the distribution of the test statistic under the null hypothesis?

6. Obtain the p-value (or a range if using Table 4).

7. Using a significance level of 0.05, state your conclusions.

C. Repeat the questions in (b) for the bean diet.

D. Now let’s find out if one diet performs better than the other.

1. What kind of sample do we have? Two independent, single or paired?

2. What kind of question are we asking? Two-sided/nondirectional or one- sided/directional?

3. State the null and alternative hypotheses.

4. Write out and calculate the t-statistic.

5. What is the distribution of the test statistic under the null hypothesis?

6. Either by hand or using the R code on page 1, draw the null distribution and indicate the value of the t-statistic using a vertical line. Shade or indicate the area under the curve corresponding to the p-value.

7. Using R or Table 4, obtain the p-value (or a range if using Table 4). In R, pt(q, df) returns Pr(t<q) for a t distribution with df degrees of freedom.7.

8. Using a significance level of 0.05, state your conclusions.

e. Confidence intervals (CIs)

1. Construct a 95% CI for the mean fall in cholesterol under the bean diet.

2. Construct a 95% CI for the mean fall in cholesterol under the oat diet.

3. Construct a 95% CI for the mean difference between the two diets (bean – oat).

4. Does your conclusion from part (d) agree with your CI for the difference between

the two diets? Explain why this should or should not be expected.

Explanation / Answer

Q1.
a.
Standard Error= sd/ Sqrt(n)
Where,
sd = Standard Deviation
n = Sample Size
Standard deviation( sd )=31.1
Sample Size(n)=10
Standard Error = ( 31.1/ Sqrt ( 10) )
= 9.83
b.
Standard Error= sd/ Sqrt(n)
Where,
sd = Standard Deviation
n = Sample Size
Standard deviation( sd )=29.4
Sample Size(n)=10
Standard Error = ( 29.4/ Sqrt ( 10) )
= 9.3
Q2.
a.
Standard Error= sd/ Sqrt(n)
Where,
sd = Standard Deviation
n = Sample Size
Standard deviation( sd )=31.1
Sample Size(n)=50
Standard Error = ( 31.1/ Sqrt ( 50) )
= 4.4
b.
Standard Error= sd/ Sqrt(n)
Where,
sd = Standard Deviation
n = Sample Size
Standard deviation( sd )=29.4
Sample Size(n)=50
Standard Error = ( 29.4/ Sqrt ( 50) )
= 4.16
Q3.
Given that,
mean(x)=53.6
standard deviation , s.d1=31.1
number(n1)=10
y(mean)=55.5
standard deviation, s.d2 =29.4
number(n2)=10
null, Ho: u1 = u2
alternate,diet reduces cholesterol in this population on average H1: u1 < u2
level of significance, = 0.05
from standard normal table,left tailed t /2 =1.833
since our test is left-tailed
reject Ho, if to < -1.833
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 1.833
we got |to| = 0.14039 & | t | = 1.833
make decision
hence value of |to | < | t | and here we do not reject Ho
p-value:left tail - Ha : ( p < -0.1404 ) = 0.44572
hence value of p0.05 < 0.44572,here we do not reject Ho
ANSWERS
---------------
null, Ho: u1 > u2
alternate, H1: u1 < u2
test statistic: -0.14
critical value: -1.833
decision: do not reject Ho
p-value: 0.44572

we don't have evidence to support diet reduces cholesterol in this population on average

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