Statistics And Regression Analysis

Question 1: Weight Reduction Program Statistical Analysis

Question: 1

Let X is weight losses, in pounds.

Assumption: X follows normal distribution.

a)

Weight Loss Save Time On Research and Writing Hire a Pro to Write You a 100% Plagiarism-Free Paper. Get My Paper
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
p-value	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)

F-test 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.  =

Question 2: Matched Pairs Experiment Analysis

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.

H₀:  =

vs

H₁:   

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

H₀:  vs H₁:

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 H₀  follows t distribution with  degrees of freedom.

Decision criteria:

Reject H₀  if

From given data:

MP-Sam-1	MP-Sam-2	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.

Question 3: Stock Price and Interest Rate Regression Analysis

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 :

	Sample-1	Sample-2
Mean	59.83	50.25
Variance	415.7879	401.8409
Observations	12	12
Pooled Variance	408.8144

For paired t test:

	MP-Sam-1	MP-Sam-2
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 t-test 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 H₀:  vs H₁:

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  H₀ 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 .

Reference:

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. 269-278). 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.