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Statistics II 2077 question paper

The complete TU 2077 exam paper for Statistics II (STA215), all 12 questions with solved model answers written to the mark scheme.

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110 marksNumericalsampling distribution of mean and proportiHideAnswer
Describe the concept of sampling distribution of mean with reference to the population data (20, 21, 22 & 23) of size 4. In order to explain this, perform simple random sampling with replacement taking all possible samples with sample size n = 2. While describing the sampling distribution following issues will be covered: a. population mean & population variance, and its distribution b. Sample mean & sample variance, and its distribution c. Comparison of population mean and sample mean; population variance and sample variance; population distribution and sampling distribution based on the given data. d. Standard error of mean e. Final comments based on your result[10]
Population values: $X = \{20, 21, 22, 23\}$ - Population size: $N = 4$ - Sampling: Simple Random Sampling With Replacement (SRSWR) - Sample size: $n = 2$ - All possible samples taken All required data present. --- $$\mu = \frac{\sum X}...

Population values: $X = {20, 21, 22, 23}$ - Population size: $N = 4$ - Sampling: Simple Random Sampling With Replacement (SRSWR) - Sample size: $n = 2$ - All possible samples taken All required data present. --- $$\mu = \frac{\sum X}...
210 marksNumericalIntroduction of multiple linear regressionHideAnswer
Multiple Linear Regression Analysis
Obs Lifetime (years) Play time (hrs/day) RAM (MB) ------------------------------------------------------ 1 5 2 8 2 1 8 2 3 7 1 6 4 2 5 3 5 3 6 2 6 4 3 4 7 6 2 7 $n = 7$ . Dependent variable: Lifetime (Y), since it is being affected/explai...
Obs Lifetime (years) Play time (hrs/day) RAM (MB) ------------------------------------------------------ 1 5 2 8 2 1 8 2 3 7 1 6 4 2 5 3 5 3 6 2 6 4 3 4 7 6 2 7 $n = 7$. Dependent variable: Lifetime (Y), since it is being affected/explai...
310 marksNumericalLatin Square DesignHideAnswer
Latin Square Design (LSD) - Concepts, Conditions, and Analysis
Definition: Latin Square Design is an experimental design used to control variability in two directions (rows and columns) simultaneously. The material is heterogeneous in two perpendicular directions, and local control is applied in bot...
45 marksNumericalone sample tests for mean of normal populaHideAnswer
A dealer of a DELL company located at New Road claimed that the average lifetime of a multimedia projector produced by Dell Company is greater than 60,000 hours with standard deviation of 6000 hours. In order to test his claim, sample of 100 DELL projectors are taken and the average life time was monitored and it was found to be 55,000 hours. Test the claim of the dealer at 5% level of significance. [5]
Parameter Value ------------------ Claimed population mean $\mu = 60{,}000$ hours Population standard deviation $\sigma = 6{,}000$ hours Sample size $n = 100$ Sample mean $\bar{X} = 55{,}000$ hours Level of significance $\alpha = 0.05$ S...
Parameter Value ------------------ Claimed population mean $\mu = 60{,}000$ hours Population standard deviation $\sigma = 6{,}000$ hours Sample size $n = 100$ Sample mean $\bar{X} = 55{,}000$ hours Level of significance $\alpha = 0.05$ S...
55 marksNumericalLinkage between confidence interval and teHideAnswer
Independent t-test and Confidence Interval Analysis

Based on the following information, perform the following:

a. Test whether two means are significantly different (α = 5%) using independent t-test.

b. Compare 95% confidence interval estimation for the difference of means.

c. Show the linkage between testing of hypothesis and confidence interval estimation in this problem.

Group I Group B
Sample mean 10 15
Sample Standard Deviation 3 5
Sample Size 49 64

[5]
Group I Group B --------- Sample mean $\bar{X}1 = 10$ $\bar{X}2 = 15$ Sample SD $S1 = 3$ $S2 = 5$ Sample size $n1 = 49$ $n2 = 64$ - Significance level: $\alpha = 0.05$ - Confidence level: 95% Note: Both samples are large ( $n1, n2 30$ ), s...
Group I Group B --------- Sample mean $\bar{X}1 = 10$ $\bar{X}2 = 15$ Sample SD $S1 = 3$ $S2 = 5$ Sample size $n1 = 49$ $n2 = 64$ - Significance level: $\alpha = 0.05$ - Confidence level: 95% Note: Both samples are large ($n1, n2 30$), s...
65 marksNumericalDetermination of sample sizeHideAnswer
A study of 1000 computer engineers conducted by their professional organization reported that 300 stated that their firms’ greatest concern was to uplift the professional quality of work. In order to conduct a follow up study to estimate the population proportion of computer engineers to fulfill their greatest concern within ±0.01 with 99% confidence interval, how many computer engineers would be required to be surveyed? [5]
Parameter Value ------------------ Preliminary sample size $n0 = 1000$ Number with greatest concern $X = 300$ Sample proportion $\hat{p} = 300/1000 = 0.30$ Margin of error $E = 0.01$ Confidence level $99\%$ --- Step 1: Determine p and q ...
Parameter Value ------------------ Preliminary sample size $n0 = 1000$ Number with greatest concern $X = 300$ Sample proportion $\hat{p} = 300/1000 = 0.30$ Margin of error $E = 0.01$ Confidence level $99%$ --- Step 1: Determine p and q ...
75 marksNumericalTwo independent sample testHideAnswer
Chi-Square Test for Association Between Email Account Type and Hacking Status

A survey was conducted to see the association between hacking status of the email and the type of email account. The survey has reported the following cross tabulation. Do the information provide sufficient evidence to conclude that the type email account and the hacking status is associated? Use Chi-square test at 1% level of significance.

$\begin{array}{|c|c|c|} \hline \text{Type of e-mail account} & \text{Hacking status Yes} & \text{Hacking status No} \\ \hline \text{Yahoo} & 60 & 15 \\ \text{Gmail} & 20 & 120 \\ \hline \end{array}$

[5]
Chi-Square Test for Association Between Email Account Type and Hacking Status

A survey was conducted to see the association between hacking status of the email and the type of email account. The survey has reported the following cross tabulation. Do the information provide sufficient evidence to conclude that the type email account and the hacking status is associated? Use Chi-square test at 1% level of significance.

$$\begin{array}{|c|c|c|} \hline \text{Type of e-mail account} & \text{Hacking status Yes} & \text{Hacking status No} \ \hline \text{Yahoo} & 60 & 15 \ \text{Gmail} & 20 & 120 \ \hline \end{array}$$

[5]
Chi-Square Test for Association Between Email Account Type and Hacking Status

Step 1 - Given Data

Account Yes (Hacked) No (Not Hacked) Row Total
Yahoo 60 15 75
Gmail 20 120 140
Column Total 80 135 215

Level of significance: $\alpha = 0.01$

Step 2 - Hypotheses
- $H_0$ : Type of email account and hacking status are independent (not associated).
- $H_1$ : Type of email account and hacking status are associated.
Step 3 - Expected Frequencies

$E_{ij} = \frac{\text{Row Total} \times \text{Column Total}}{\text{Grand Total}}$

$E_{11} = \frac{75 \times 80}{215} = 27.907$

$E_{12} = \frac{75 \times 135}{215} = 47.093$

$E_{21} = \frac{140 \times 80}{215} = 52.093$

$E_{22} = \frac{140 \times 135}{215} = 87.907$

Step 4 - Chi-Square Statistic

$\chi^2 = \sum \frac{(O-E)^2}{E}$

O E $(O-E)$ $(O-E)^2$ $(O-E)^2/E$
60 27.907 32.093 1029.96 36.906
15 47.093 -32.093 1029.96 21.870
20 52.093 -32.093 1029.96 19.771
120 87.907 32.093 1029.96 11.717

$\chi^2 = 36.906 + 21.870 + 19.771 + 11.717 = 90.264 \approx 90.26$

Step 5 - Degrees of Freedom

$df = (r-1)(c-1) = (2-1)(2-1) = 1$

Step 6 - Critical Value

$\chi^2_{tab,\,0.01,\,1} = 6.635$

Step 7 - Decision

Since $\chi^2_{cal} = 90.26 > \chi^2_{tab} = 6.635$ , we reject $H_0$ .

Conclusion

At the 1% level of significance, there is sufficient evidence to conclude that the type of email account and hacking status are significantly associated. (Yahoo accounts show a much higher hacking rate than Gmail accounts.)

Note: For a 2×2 table Yates' continuity correction is sometimes applied, giving $\chi^2 \approx 87.6$ , but the conclusion (reject $H_0$ ) remains unchanged.
Chi-Square Test for Association Between Email Account Type and Hacking Status

Step 1 - Given Data

Account Yes (Hacked) No (Not Hacked) Row Total
Yahoo 60 15 75
Gmail 20 120 140
Column Total 80 135 215

Level of significance: $\alpha = 0.01$

Step 2 - Hypotheses
- $H_0$: Type of email account and hacking status are independent (not associated).
- $H_1$: Type of email account and hacking status are associated.
Step 3 - Expected Frequencies

$$E_{ij} = \frac{\text{Row Total} \times \text{Column Total}}{\text{Grand Total}}$$

$$E_{11} = \frac{75 \times 80}{215} = 27.907$$

$$E_{12} = \frac{75 \times 135}{215} = 47.093$$

$$E_{21} = \frac{140 \times 80}{215} = 52.093$$

$$E_{22} = \frac{140 \times 135}{215} = 87.907$$

Step 4 - Chi-Square Statistic

$$\chi^2 = \sum \frac{(O-E)^2}{E}$$

O E $(O-E)$ $(O-E)^2$ $(O-E)^2/E$
60 27.907 32.093 1029.96 36.906
15 47.093 -32.093 1029.96 21.870
20 52.093 -32.093 1029.96 19.771
120 87.907 32.093 1029.96 11.717

$$\chi^2 = 36.906 + 21.870 + 19.771 + 11.717 = 90.264 \approx 90.26$$

Step 5 - Degrees of Freedom

$$df = (r-1)(c-1) = (2-1)(2-1) = 1$$

Step 6 - Critical Value

$$\chi^2_{tab,,0.01,,1} = 6.635$$

Step 7 - Decision

Since $\chi^2_{cal} = 90.26 > \chi^2_{tab} = 6.635$, we reject $H_0$.

Conclusion

At the 1% level of significance, there is sufficient evidence to conclude that the type of email account and hacking status are significantly associated. (Yahoo accounts show a much higher hacking rate than Gmail accounts.)

Note: For a 2×2 table Yates' continuity correction is sometimes applied, giving $\chi^2 \approx 87.6$, but the conclusion (reject $H_0$) remains unchanged.
85 marksNumericalEstimationHideAnswer
A machine produce metal rods used in an automobile suspension system. A random sample of 6 rods is selected and diameter is measured. The measuring data (in millimeters) are as follows. Assuming that the sample drawn from the normally distributed population. Find 95% two sided confidence interval on the mean rod diameter, and interpret the result with reference to the given problem.

8.24 8.26 8.20 8.28 8.21 8.23

[5]
Sample size: $n = 6$ - Data (mm): $8.24, 8.26, 8.20, 8.28, 8.21, 8.23$ - Population: normally distributed - Population standard deviation $\sigma$ : unknown - Confidence level: $95\%$ , so $\alpha = 0.05$ Since $\sigma$ is unknown and

Sample size: $n = 6$ - Data (mm): $8.24, 8.26, 8.20, 8.28, 8.21, 8.23$ - Population: normally distributed - Population standard deviation $\sigma$: unknown - Confidence level: $95%$, so $\alpha = 0.05$ Since $\sigma$ is unknown and
95 marksNumericalTwo independent sample testHideAnswer
Use Mann-Whitney U test to assess whether the following satisfaction scores based on the performance of two different special types of gadgets at 5% level of significance.

1 2 3 4
Gadget A 50 40 30 20
Gadget B 40 30 10 40

[5]
Gadget A (n₁ = 4): 50, 40, 30, 20 Gadget B (n₂ = 4): 40, 30, 10, 40 Level of significance: α = 0.05 - H₀: No significant difference in satisfaction scores between Gadget A and Gadget B. - H₁: There is a significant difference (two-tailed...
105 marksMarkov ProcessHideAnswer
Define Markov chain and describe its characteristics. [5]
A Markov Chain is a special type of stochastic process in which the future state of the system depends only on the present state, and not on the past states (history). It is a discrete-time stochastic process {X(t), t = 0, 1, 2, ...} tha...
115 marksNumericalMarkov ProcessHideAnswer
Every day is generally considered as either sunny or rainy. A sunny day is followed by another sunny day with probability 0.8 whereas a rainy day is followed by a sunny day with probability 0.4. Suppose it rains on Monday. Make forecasts for Tuesday and Wednesday. [5]
Given data: - States: Sunny (S), Rainy (R) - P(Sunny Sunny) = 0.8, therefore P(Rainy Sunny) = 0.2 - P(Sunny Rainy) = 0.4, therefore P(Rainy Rainy) = 0.6 - Initial condition: it rains on Monday, so P(Rainy on Monday) = 1 - Required: forec...
125 marksModel adequacy testsHideAnswer
Write shorts notes on the following: i. Test of equality of two variances ii. Adjusted R2R^2R2 [5]
The test of equality of two variances (also called the F-test) is used to determine whether two population variances are equal. It is based on the F-distribution and is a parametric test. - Null Hypothesis (H₀): σ₁² = σ₂² (The two popula...

	Group I	Group B
Sample mean	10	15
Sample Standard Deviation	3	5
Sample Size	49	64

Account	Yes (Hacked)	No (Not Hacked)	Row Total
Yahoo	60	15	75
Gmail	20	120	140
Column Total	80	135	215

O	E	$(O-E)$	$(O-E)^2$	$(O-E)^2/E$
60	27.907	32.093	1029.96	36.906
15	47.093	-32.093	1029.96	21.870
20	52.093	-32.093	1029.96	19.771
120	87.907	32.093	1029.96	11.717

Account	Yes (Hacked)	No (Not Hacked)	Row Total
Yahoo	60	15	75
Gmail	20	120	140
Column Total	80	135	215

O	E	$(O-E)$	$(O-E)^2$	$(O-E)^2/E$
60	27.907	32.093	1029.96	36.906
15	47.093	-32.093	1029.96	21.870
20	52.093	-32.093	1029.96	19.771
120	87.907	32.093	1029.96	11.717


8.24	8.26	8.20	8.28	8.21	8.23

	1	2	3	4
Gadget A	50	40	30	20
Gadget B	40	30	10	40

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Statistics II 2077 question paper

Chi-Square Test for Association Between Email Account Type and Hacking Status

Step 1 - Given Data

Step 2 - Hypotheses

Step 3 - Expected Frequencies

Step 4 - Chi-Square Statistic

Step 5 - Degrees of Freedom

Step 6 - Critical Value

Step 7 - Decision

Conclusion

Chi-Square Test for Association Between Email Account Type and Hacking Status

Step 1 - Given Data

Step 2 - Hypotheses

Step 3 - Expected Frequencies

Step 4 - Chi-Square Statistic

Step 5 - Degrees of Freedom

Step 6 - Critical Value

Step 7 - Decision

Conclusion

Chi-Square Test for Association Between Email Account Type and Hacking Status

Step 1 - Given Data

Step 2 - Hypotheses

Step 3 - Expected Frequencies

Step 4 - Chi-Square Statistic

Step 5 - Degrees of Freedom

Step 6 - Critical Value

Step 7 - Decision

Conclusion