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This article discusses various statistical tests and their interpretations, including confidence intervals, hypothesis testing, F-test, t-test, and regression analysis. It also covers the relationship between interest rates and stock prices, and the significance of Treasury bills as a predictor of SP 500. The article concludes with a brief overview of Desklib, an online library for study material with solved assignments, essays, dissertations, etc., covering a wide range of subjects, courses, and universities.

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.
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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 = ( X −20 ¿ / (s/ √ n)
Where X = mean of X = 16.37
s = 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)
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p value = P(T <T Cal)
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=( X −20 ¿ / (5/ √ n)
So,
Z < -1.64
Is equivalent to X < -1.64 × (5/ √n) + 20
i.e. X < -1.64 × (5/ √ 10) + 20
X < 17.40 where X follows normal distribution.
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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 β.
So, β = P( X ≥ 17.40 when X follows normal distribution with mean 19 and s. d. 5/ √10)
β = P(( X -19)/ (5 / √ 10) ≥ (17.40 - 19)/ (5 / √ 10) )
β = P( Z ≥-1.0124 )
β = 0.844
Question 2;
a)
Provided in Excel Sheet
b)
F-test for comparing two variances:
Suppose σ 1
2 and σ 2
2 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. σ 1
2 = σ 2
2
Vs
Alternative Hypothesis: Both the population variance differ from each other i.e. σ 1
2 ≠ σ 2
2.
Test statistic for testing the above null and alternative hypothesis is
F Calculated = S1
2 / S2
2
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Under null hypothesis, F calculated follows F Distribution with (n 1−1) and (n 2−2) degrees
of freedom. Where S1
2 is the sample variance of sample1 and S2
2 is the sample variance of
sample2.
Decision criteria:
We reject the null hypothesis if F Calculated > F(1− α
2 ,n 1−1, n 2−1) or F Calculated < F( α
2 ,n 1−1 ,n 2−1)
F( α
2 ,n 1−1 ,n 2−1) and F(1− α
2 ,n 1−1, n 2−1) are critical values of F distribution.
For given sample 1 and sample 2:
S1
2 = 415.79 and S2
2 = 401.84
n1= 12 and n2 = 12
So
F Calculated = S1
2 / S2
2 = 415.79 / 401.84 = 1.03471531
At α=0.05, critical values are F(0.025 ,11,11)=¿ 0.2879 and F(0.975,11,11)=¿ 3.4737
So,
F(0.025 ,11,11)=¿ 0.2879 < F Calculated = 1.03471531 < F(0.975,11,11)=¿ 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 X1 and X2. Let μ1 and μ2 are the population mean of sample 1 and sample 2
rspectively.
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H0: μ1 = μ2
vs
H1: μ1 ≠ μ2
For testing the above hypothesis test statistics is
t=( X1−X2 )−( μ1−μ2 )
S √ 1
n 1 + 1
n2
n 1 size of sample1, n 2 is size of sample 2,
S2 is pooled variance which is defined as
S2= ( n 1−1 ) × S1
2 + ( n 2−1 ) × S2
2
n1+ n2−2
Where S1
2 is the sample variance of sample1 and
S2
2 is the sample variance of sample2.
Under the null hypothesis, t follows t distribution with (n 1+n 2−2) degrees of freedom.
Decision criteria:
We reject the H0 if |t |>t(1− α
2 ,n 1+n 2−2)
For given sample 1 and sample 2:
S1
2 = 415.79 and S2
2 = 401.84
n1= 12 and n2 = 12
So S=408.81
X1= 59.83 and X2 = 50.25
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So, test statistics t= 1.161
Critical value:
t(1− α
2 ,n 1+n 2−2) = 2.074
So, we fail to reject the null hypothesis as
|t=1.161|<t(1− α
2 ,n 1+ n 2−2 )=2.074
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 di= X1 i−X2 i
H0: μd =0 vs H1: μd ≠ 0
Where μd population mean of di.
Test statistics for testing the null hypothesis vs alternative hypothesis is
t= d−μd
Sd / √ n
Where d is mean of di
Sd Sample s. d. of di
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n is the number of pairs
Under H0 t follows t distribution with (n−1) degrees of freedom.
Decision criteria:
Reject H0 if |t |>t(1− α
2 ,n−1)
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.
e)
Yes, required condition satisfied.
First condition is Independence for two independent sample t test and dependence for paired t
test.
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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
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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.
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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: β2=0 vs H1: β2 ≠ 0
Test statistics t for testing above hypothesis is
t= ^β2
SE ( ^β2 ) =−99.4014
22.67665 =−4.38343
Under Null hypothesis, t follows t distribution with 26 degrees of freedom.
Critical values of t26 at α = 0.05 is
Critical t value = 2.056
Decision criteria for rejecting null hypothesis:
Reject H0 if |t |>Critical t Value
So |-4.38343| > 2.056
So we reject Null hypothesis. i.e. slope coefficient is significant at 5%.
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Question 4:
Given ^β1=
∑
i=1
n
( Xi− X ) (Y i −Y )
∑
i=1
n
( Xi −X )2
Consider,
∑
i=1
n
( Xi− X ) ( Y i−Y ) =∑
i=1
n
( Xi Y i −X Y i−Y Xi +X ¿Y )=∑
i=1
n
Xi Y i −X ∑
i=1
n
Y i−Y ∑
i=1
n
X i+ n X Y ¿=∑
i=1
n
Xi Y i−n X Y −n
So we can write ^β1 as
^β1=a+b1 Y 1+b2 Y 2 +…+ bn Y n-------------------------------(1)
Where a= −n X Y
∑
i=1
n
( Xi− X )2 , b1= X1
∑
i=1
n
( Xi− X )2 , b2= X2
∑
i =1
n
( Xi− X )2 , … . bn= Xn
∑
i=1
n
( Xi− X )2 ,
We can observe that equation (1) is the linear function of response variable Y i.
Reference:
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Bickel, P.J. and Doksum, K.A., 2015. Mathematical statistics: basic ideas and selected
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Chatterjee, S. and Hadi, A.S., 2015. Regression analysis by example. John Wiley & Sons.
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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.
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