Question 1
A clinic offers a weight reduction program. A review of its records found the following weight losses, in pounds, for a random sample of ten of its patients at the conclusion of the program.
18.2 25.9
6.3
11.8 15.4
20.3 16.8 19.5 12.3 17.2
Assume the population distribution is normal. Use Excel to compute necessary statistics and probabilities.
(a) Find a 99% confidence interval for the population mean weightloss.
(b) Using the result in (a), test at the 1% significance level the null hypothesis that the population mean equals 20 against the twosided alternative.
(c) Test the null hypothesis that the population mean weightloss equals 20 pounds against the onesided alternative hypothesis that the population mean weightloss is lower than 20 pounds at the 5% level.
(d) Find the pvalue of the above test in (c).
(e) Suppose that the population mean is in fact 20. What is theprobability that the null hypothesis will be rejected (Type I error) in (a)?
(f) Suppose the population mean is in fact 19 and the population standard deviation is 5. What is the probability that the test
H0: ???? = 20,
H1: ???? < 20
will not reject the null against the alternative at the 5% level (Type II error)?
Question 2
Samples of size 12 were drawn independently from two normal populations. A matched pairs experiment was then conducted from the same populations. These data are stored in Excel file named “Two_samples.xls”.
differences occurred.
Question 3
Stock prices and interest rates are key economic indicators. Investors in stock markets,individual or institutional, watch very carefully the movements in the interest rate.As a measure of stock prices, let us use the S&P 500 composite index, and as a measure of the interest rate, let us use the threemonth Treasury bill rate. The data on these variables are provided in Stock.xlsx for the period 19802007.
a) What relationship, if any, do you expect between interest rates and stock prices? Why?
b) Plot using Eviews or Excel the scattergram between interest rates (x axis) and stock prices (yaxis). Does the scattergram support your expectation in
part (a)?
c) Let ???????? be the S&P500 index and ???????? be the threemonth Treasury bill rate.Using Eviews or Excel, estimate the following regression
Provide the Eviews or Excel output and write down the fitted equation (include the predictive equation, tstatistics of the estimated model parameters, and sample size).0 1
d) Interpret the coefficients. Does the sign of the slope coefficientconsistent with your expectation?Question 4
Suppose the regression model is
The OLS estimator of the slope coefficient is
Show that the OLS slope estimator is a linear fu
Question: 1
Let X is weight losses, in pounds.
Assumption: X follows normal distribution.
a)
Weight Loss 

Mean 
16.37 
Standard Error 
1.699938 
Median 
17 
Mode 
#N/A 
Standard Deviation 
5.375676 
Sample Variance 
28.89789 
Kurtosis 
0.780412 
Skewness 
0.19782 
Range 
19.6 
Minimum 
6.3 
Maximum 
25.9 
Sum 
163.7 
Count 
10 
Confidence Level (99.0%) 
5.524519 
Lower Confidence Limit = Mean  Confidence Level (99.0%) = 16.37 – 5.524519 = 10.84548
Upper Confidence Limit = Mean + Confidence Level (99.0%) = 16.37 + 5.524519 = 21.89452
So, 99% Confidence interval is (10.84548, 21.89452)
b)
As 99% Confidence interval includes the value 20. We accept null hypothesis.
c)
Here we test the following hypothesis:
Null Hypothesis: Population mean weight loss is 20 pounds.
Vs
Alternative hypothesis: Population mean weight loss is lower than 20 pounds.
So test statistic for testing this hypothesis is
T cal = ( /
Where = mean of X = 16.37
= standard deviation of X = 5.3757
n = number of observation = 10
So T cal = 2.14
Decision criteria:
Reject the null hypothesis if T cal < Ctitical T Value.
Critical T Value at 0.01 significance level = 1.83
So, T Cal < Critical T Value, so we reject null hypothesis.
Mean 
16.37 
SD 
5.375676 
n 
10 
T Cal 
2.13537 
Critical T Value 
1.83311 
d)
p value =
where T has t distribution with (n  1) = 9 degrees of freedom.
p value = P( T < 2.14) = 0.031
Excel Output:
n 
10 
T Cal 
2.13537 
Critical T Value 
1.83311 
pvalue 
0.030741 
e)
Type I error is nothing but error occurred by rejecting the null hypothesis when it is true.
Significance level is the the probability of type I error.
In c) we have 0.05 is the significance level. So, Probability of type I error is 0.05.
f)
For the testing given null hypothesis against the alternative hypothesis, we reject the null hypothesis if Z < 1.64
Where Z=( /
So,
Z < 1.64
Is equivalent to < 1.64 × + 20
i.e. < 1.64 × + 20
< 17.40 where follows normal distribution.
Probability of Type II error:
Probability of type error is the probability of reject alternative hypothesis when alternative hypothesis is true. It is denoted by .
So, = P( 17.40 when follows normal distribution with mean 19 and s. d. 5/)
= P((19)/ (5 / ) (17.40  19)/ (5 / ) )
= P( Z 1.0124 )
= 0.844
Question 2;
a)
Provided in Excel Sheet
b)
Ftest for comparing two variances:
Suppose and are population variance for sample1 and sample2 respectively. Here we test whether the two population variances are equal or not. We formulate the following null and alternative hypothesis.
Null Hypothesis: Both the population have equal variances i.e. =
Vs
Alternative Hypothesis: Both the population variance differ from each other i.e. .
Test statistic for testing the above null and alternative hypothesis is
F Calculated = /
Under null hypothesis, F calculated follows F Distribution with and degrees of freedom. Where is the sample variance of sample1 and is the sample variance of sample2.
Decision criteria:
We reject the null hypothesis if F Calculated > or F Calculated <
and are critical values of F distribution.
For given sample 1 and sample 2:
= 415.79 and = 401.84
n1= 12 and n2 = 12
So
F Calculated = / = 415.79 / 401.84 = 1.03471531
At , critical values are 0.2879 and 3.4737
So,
0.2879 < F Calculated = 1.03471531 < 3.4737
We accept the null hypothesis. i.e. both the population has same variances.
c)
To test the null hypothesis of the mean of the two populations are equal, based on two random samples. That is, to investigate the significance of the difference between the two sample means and . Let and are the population mean of sample 1 and sample 2 rspectively.
H_{0}: =
vs
H_{1}:
For testing the above hypothesis test statistics is
size of sample1, is size of sample 2,
is pooled variance which is defined as
Where is the sample variance of sample1 and
is the sample variance of sample2.
Under the null hypothesis, t follows t distribution with degrees of freedom.
Decision criteria:
We reject the if
For given sample 1 and sample 2:
= 415.79 and = 401.84
n1= 12 and n2 = 12
So
= 59.83 and = 50.25
So, test statistics t= 1.161
Critical value:
= 2.074
So, we fail to reject the null hypothesis as
Means of two population from which sample1 and sample2 is drawn have same mean.
d)
In matched pair data,
We are interested in testing the null hypothesis: There is no any difference before and after against
Alternative hypothesis: There is significant difference between before and after.
We define
H_{0}: vs H_{1}:
Where population mean of .
Test statistics for testing the null hypothesis vs alternative hypothesis is
Where is mean of
Sample s. d. of
is the number of pairs
Under H_{0} follows t distribution with degrees of freedom.
Decision criteria:
Reject H_{0} if
From given data:
MPSam1 
MPSam2 
di 
55 
48 
7 
45 
37 
8 
52 
43 
9 
87 
75 
12 
78 
78 
0 
42 
35 
7 
62 
45 
17 
90 
79 
11 
23 
12 
11 
60 
53 
7 
67 
59 
8 
53 
37 
16 

Sum 
113 

Mean 
9.416667 

n 
12 

Sd 
4.501683 

t 
7.246243 

alpha 
0.05 

Critical Value 
2.200985 
So we reject null hypothesis. There is significant difference between means of matched pair samples.
e)
Yes, required condition satisfied.
First condition is Independence for two independent sample t test and dependence for paired t test.
For independent two sample t test, two samples must be drawn from randomly and independently.
For Paired t test data is dependent.
As in c) we have two different sample drawn from different population where as in d) measurement are taken from same unit two times.
Randomization is second condition which is also satisfied as they are selected randomly from the population. This is condition for both two independent sample t test and paired t test.
Third condition is normality. As sample size is less than 30 we used t distribution as our data does not have any outlies.
f)
We observed the following mean for two independent sample t test and paired t test as
For two independent sample t test :

Sample1 
Sample2 
Mean 
59.83 
50.25 
Variance 
415.7879 
401.8409 
Observations 
12 
12 
Pooled Variance 
408.8144 

For paired t test:

MPSam1 
MPSam2 
Mean 
59.5 
50.08333 
Variance 
369 
402.2652 
We observed that there is very little change between difference means of samples for two independent sample t test and paired t test. But the degrees of freedom for two independent sample t test is 22 and for paired t test is 11.
Two independent sample ttest is used when we compare means from two different populations whereas paired t test is used when data is (dependent) collected from same unit two times (before and after type)
Question 3:
a)
When interest rate go up, stock prices goes down.
As interest rate goes up, people deposit their money in the bank than investing in stock. So demand of stock decreases as price decreases.
b)
We can see that there is negative relationship between SP 500 and Treasury Bills. So this scatter plot support our expectation. As when Treasury bill increases SP 500 decreases and when T Bill decreases SP 500 increases.
c)
Here we fit the simple regression model to the SP 500. We used Treasury bill as predictor. Following output shows the result of fitting the regression model to SP 500. This output also gives the significance test for both the coefficient intercept and slope. The model fitting is not very good as we can observe R square is only .424962. i.e. out of the total variation in SP 500, 42.49% variation is explained by Treasury bill.
Regression Equation:
SP500 = 1229.341 – 99.4014 × Tbill
d)
Interpretation of coefficient:
Intercept:
When Tbill is zero then SP500 = 1229.341
Slope:
When Tbill increases by unit then SP500 decreases by 99.4014
e)
Here we test H_{0}: vs H_{1}:
Test statistics t for testing above hypothesis is
Under Null hypothesis, t follows t distribution with 26 degrees of freedom.
Critical values of at = 0.05 is
Critical t value = 2.056
Decision criteria for rejecting null hypothesis:
Reject H_{0} if
So 4.38343 > 2.056
So we reject Null hypothesis. i.e. slope coefficient is significant at 5%.
Question 4:
Given
Consider,
So we can write as
(1)
Where , , , … . ,
We can observe that equation (1) is the linear function of response variable .
Bickel, P.J. and Doksum, K.A., 2015. Mathematical statistics: basic ideas and selected topics, volume I (Vol. 117). CRC Press.
Chatterjee, S. and Hadi, A.S., 2015. Regression analysis by example. John Wiley & Sons.
DeGroot, M.H. and Schervish, M.J., 2012. Probability and statistics. Pearson Education.
Draper, N.R. and Smith, H., 2014. Applied regression analysis (Vol. 326). John Wiley & Sons.
Hogg, R.V. and Craig, A.T., 1995. Introduction to mathematical statistics.(5"" edition) (pp. 269278). Upper Saddle River, New Jersey: Prentice Hall.
Moyé, L. A., Chan, W., & Kapadia, A. S. (2017). Mathematical statistics with applications. CRC Press.
Ross, S.M., 2014. Introduction to probability and statistics for engineers and scientists. Academic Press.
Ryan, T.P., 2008. Modern regression methods (Vol. 655). John Wiley & Sons.
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