IGNOU MST-016 Solved Assignment 2024 Cover Page

IGNOU MST-016 Solved Assignment 2024 | MSCAST | IGNOU

Solved By – Narendra Kr. Sharma – M.Sc (Mathematics Honors) – Delhi University

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IGNOU MST-016 Assignment Question Paper 2024

mst-016-solved-assignment-2024-qp-8e24e610-06c9-4b43-84f6-a5bf6ef5ab5c

(a) State whether the following statements are True or False. Give reason in support of your answer:
(i) If $X_{1}, X_{2}, X_{3}, X_{4}$ and $X_{5}$ is a random sample of size 5 taken from an Exponential distribution, then estimator $T_{1}$ is more efficient than $T_{2}$ .

T_{1} = \frac{X_{1} + X_{2} + X_{3} + X_{4} + X_{5}}{5}, T_{2} = \frac{X_{1} + 2 X_{2} + 3 X_{3} + 4 X_{4} + 5 X_{5}}{15}

(ii) If

T_{1}

and

T_{2}

are two estimators of the parameter

θ

such that

Var (T_{1}) = 1 / n

and

Var (T_{2}) = n

then

T_{1}

is more efficient than

T_{2}

.
(iii) A

95 %

confidence interval is smaller than

99 %

confidence interval.
(iv) If the probability density function of a random variable

X

follows

F

-distribution is

f (x) = \frac{1}{(1 + x)^{2}}, x \geq 0

then degrees of freedom of the distribution will be

(2, 2)

.
(v) A patient suffering from fever reaches to a doctor and suppose the doctor formulate the hypotheses as

H_{0}

: The patient is a chikunguniya patient

H_{1}

: The patient is not a chikunguniya patient
If the doctor rejects

H_{0}

when the patient is actually a chikunguniya patient, then the doctor commits type II error.

(b) Describe the various forms of the sampling distribution of ratio of two sample variances.
2(a) A baby-sister has 6 children under her supervision. The age of each child is as follows:

Child	Age (in years)
Sonu	2
Lavnik	4
Chiya	3
Amam	3
Avishi	4
Ridhi	5
Sidhi	3

(i) What is the form of population of age of children?
(ii) Prepare the sampling distribution of sample mean when sample size is 2.
(iii) Is the shape of the sampling distribution normal?
(iv) Calculate the mean and standard error of the sampling distribution.

The department of transportation has mandated that the average speed of cars on interstate highways be no more than $70 k m$ per hours in order. To check that the people follow it or not, a researcher took a random sample of 186 cars and found that the average speed was $72 k m$ per hours with a standard deviation $0.6 k m$ per hours.
(a) Construct the interval around the sample mean that would contain the population mean $95 %$ of the time.
(b) If the researcher wants to test that the true mean speed on its highways is $70 k m$ per hours or less with $95 %$ confidence then
(i) State null and alternative hypotheses.
(ii) Name the test which is suitable in this situation and why?
(iii) Calculate the value of test statistic and critical value.
(iv) Draw the conclusion on the basis of the applied test.
4(a) A sample of 500 shops was selected in a large metropolitan area to determine various information concerning consumer behaviour. One question, among the questions, asked, was "Do you enjoy shopping for clothing?" Out of 240 males 136 answered yes. Out of 260 females, 224 answered yes. Find $95 %$ confidence interval for the difference of the proportions for enjoys shopping for clothing.
(b) An engineer conducted an experiment to compare two metals: iron and copper, as bonding agents for an alloy material. Components of the alloy were bonded using the metals as bonding agents, and the pressures required to break the bonds were measured. The data for the breaking pressures are given in the following table:

S.	Breaking Pressure
S.	Iron	Copper
1	72.7	73
2	69.6	67.2
3	83.4	75.3
4	78.9	61.4
5	75	74
6	71.6	69.5
7	85.7	69.8
8	73.5	73.8
9	70.4	68
10	84.2	76.1

If the breaking pressures for both iron and copper are normally distributed, are the variances of the distributions of the breaking pressure of iron and copper equal at

5 %

level of significance?

\begin{aligned} 5. (a) Complete the following table, one is done for you: \\ \begin{array}{cccc} \begin{array}{l} S. \\ No \end{array} & Test For & \begin{matrix} Name of the \\ Test \end{matrix} & Test Statistic \\ 1 & \begin{array}{l} Population mean when \\ population variance is \\ known and population is \\ normal \end{array} & Z-test & Z = \frac{\bar{X} - μ}{σ / \sqrt{n}} \\ 2 & \begin{array}{l} Population mean when \\ population variance is \\ unknown and population is \\ normal \end{array} \\ 3 & \begin{array}{l} Difference of two \\ population means when \\ samples are paired, and \\ population of differences \\ follows normal distribution. \end{array} \\ 4 & \begin{array}{l} Difference of two \\ population means when \\ samples are independent, \\ and population of \\ differences follows normal \\ distribution. \end{array} \\ 5 & \begin{array}{l} Population variance when \\ the population is normal \\ distributed \end{array} \\ 6 & \begin{array}{l} Population variance when \\ the population is not \\ normal distributed \end{array} \end{array} \end{aligned}

(b) Describe the following:
(i) Curve of

F

-distribution
(ii) Mean Squared Error

MST-016 Sample Solution 2024

mst-016-solved-assignment-2024-ss-8e24e610-06c9-4b43-84f6-a5bf6ef5ab5c

(a) State whether the following statements are True or False. Give reason in support of your answer:
(i) If $X_{1}, X_{2}, X_{3}, X_{4}$ and $X_{5}$ is a random sample of size 5 taken from an Exponential distribution, then estimator $T_{1}$ is more efficient than $T_{2}$ .

T_{1} = \frac{X_{1} + X_{2} + X_{3} + X_{4} + X_{5}}{5}, T_{2} = \frac{X_{1} + 2 X_{2} + 3 X_{3} + 4 X_{4} + 5 X_{5}}{15}

Answer:

To determine which estimator is more efficient, we need to compare their variances. The estimator with the smaller variance is considered more efficient.

Let

X_{1}, X_{2}, X_{3}, X_{4}, X_{5}

be a random sample from an exponential distribution with rate parameter

λ

. The mean and variance of an exponential distribution are

\frac{1}{λ}

and

\frac{1}{λ^{2}}

respectively.

Estimator $T_{1}$

T_{1}

is the sample mean, given by:

T_{1} = \frac{X_{1} + X_{2} + X_{3} + X_{4} + X_{5}}{5}

The variance of

T_{1}

is:

Var (T_{1}) = Var (\frac{X_{1} + X_{2} + X_{3} + X_{4} + X_{5}}{5}) = \frac{1}{25} \sum_{i = 1}^{5} Var (X_{i}) = \frac{5}{25 λ^{2}} = \frac{1}{5 λ^{2}}

Estimator $T_{2}$

T_{2}

is a weighted average of the sample, given by:

T_{2} = \frac{X_{1} + 2 X_{2} + 3 X_{3} + 4 X_{4} + 5 X_{5}}{15}

The variance of

T_{2}

is:

Var (T_{2}) = Var (\frac{X_{1} + 2 X_{2} + 3 X_{3} + 4 X_{4} + 5 X_{5}}{15}) = \frac{1}{225} \sum_{i = 1}^{5} i^{2} Var (X_{i}) = \frac{1}{225 λ^{2}} \sum_{i = 1}^{5} i^{2} = \frac{55}{225 λ^{2}} = \frac{11}{45 λ^{2}}

Comparison

Comparing the variances of

T_{1}

and

T_{2}

Var (T_{1}) = \frac{1}{5 λ^{2}} < Var (T_{2}) = \frac{11}{45 λ^{2}}

Since

Var (T_{1}) < Var (T_{2})

, estimator

T_{1}

is more efficient than

T_{2}

Conclusion: The statement is true. Estimator

T_{1}

is more efficient than

T_{2}

because it has a smaller variance.

(ii) If

T_{1}

and

T_{2}

are two estimators of the parameter

θ

such that

Var (T_{1}) = 1 / n

and

Var (T_{2}) = n

then

T_{1}

is more efficient than

T_{2}

Answer:

To determine which estimator is more efficient, we compare their variances. The estimator with the smaller variance is considered more efficient.

Given:

$Var (T_{1}) = \frac{1}{n}$
$Var (T_{2}) = n$

Since

Var (T_{1}) = \frac{1}{n}

is always smaller than

Var (T_{2}) = n

for any positive

n

, we can conclude that

T_{1}

is more efficient than

T_{2}

Conclusion: The statement is true. Estimator

T_{1}

is more efficient than

T_{2}

because it has a smaller variance.

(iii) A

95 %

confidence interval is smaller than

99 %

confidence interval.

Answer:

A confidence interval (CI) gives a range of values within which the true parameter value is likely to fall, with a certain level of confidence. The width of a confidence interval depends on the level of confidence and the variability in the data. A higher confidence level means a wider interval, as it needs to cover more of the distribution to ensure that the true parameter is within the interval with higher probability.

For a normal distribution, the confidence interval is typically constructed as:

CI = \bar{x} \pm z \times \frac{s}{\sqrt{n}}

Where:

$\bar{x}$ is the sample mean,
$z$ is the z-score corresponding to the confidence level,
$s$ is the sample standard deviation,
$n$ is the sample size.

The z-score increases with the confidence level. For example:

For a $95 %$ confidence interval, $z \approx 1.96$ .
For a $99 %$ confidence interval, $z \approx 2.58$ .

Since the z-score for a

99 %

confidence interval is larger than that for a

95 %

confidence interval, the

99 %

confidence interval will be wider than the

95 %

confidence interval.

Conclusion: The statement is false. A

95 %

confidence interval is not smaller than a

99 %

confidence interval; it is actually smaller.

(iv) If the probability density function of a random variable

X

follows

F

-distribution is

f (x) = \frac{1}{(1 + x)^{2}}, x \geq 0

then degrees of freedom of the distribution will be

(2, 2)

Answer:

The given probability density function (pdf) is:

f (x) = \frac{1}{(1 + x)^{2}}, x \geq 0

This pdf does not correspond to an F-distribution. Instead, it resembles the pdf of a Pareto distribution with shape parameter

α = 2

and scale parameter

x_{m} = 1

. The F-distribution has a more complex form and depends on two degrees of freedom,

d_{1}

and

d_{2}

, which are not simply related to the parameters of this pdf.

The pdf of an F-distribution is given by:

f (x; d_{1}, d_{2}) = \frac{\sqrt{\frac{(d_{1} x)^{d_{1}} \cdot d_{2}^{d_{2}}}{(d_{1} x + d_{2})^{d_{1} + d_{2}}}}}{x \cdot B (\frac{d_{1}}{2}, \frac{d_{2}}{2})}, x \geq 0

Where

d_{1}

and

d_{2}

are the degrees of freedom, and

B

is the beta function.

Since the given pdf does not match the form of an F-distribution’s pdf, we cannot determine the degrees of freedom for the F-distribution based on the given pdf. Therefore, the statement that the degrees of freedom of the distribution are

(2, 2)

is false.

Conclusion: The statement is false. The given pdf does not correspond to an F-distribution, so we cannot determine the degrees of freedom of the distribution to be

(2, 2)

(v) A patient suffering from fever reaches to a doctor and suppose the doctor formulate the hypotheses as

H_{0}

: The patient is a chikunguniya patient

H_{1}

: The patient is not a chikunguniya patient
If the doctor rejects

H_{0}

when the patient is actually a chikunguniya patient, then the doctor commits type II error.

Answer:

In hypothesis testing, there are two types of errors:

Type I error: Occurs when the null hypothesis ( $H_{0}$ ) is rejected when it is actually true.
Type II error: Occurs when the null hypothesis ( $H_{0}$ ) is not rejected when it is actually false.

Given the hypotheses:

Null hypothesis ( $H_{0}$ ): The patient is a chikunguniya patient.
Alternative hypothesis ( $H_{1}$ ): The patient is not a chikunguniya patient.

If the doctor rejects

H_{0}

(concludes that the patient is not a chikunguniya patient) when the patient is actually a chikunguniya patient, the doctor is making a Type I error, not a Type II error.

Conclusion: The statement is false. If the doctor rejects

H_{0}

when the patient is actually a chikunguniya patient, then the doctor commits a Type I error, not a Type II error.

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IGNOU MST-016 Solved Assignment 2024 | MSCAST | IGNOU

IGNOU MST-016 Assignment Question Paper 2024

mst-016-solved-assignment-2024-qp-8e24e610-06c9-4b43-84f6-a5bf6ef5ab5c

MST-016 Sample Solution 2024