MGCR 271 Summer 2019 Assignment 3: Statistical Problem Solving

This document presents the solutions to MGCR 271 Assignment 3, focusing on statistical inference and analysis. The assignment addresses three key questions. The first question involves calculating a 99% confidence interval for a population proportion based on a survey of Tesla Model 3 owners and conducting a hypothesis test to determine if expectations were met. The second question deals with probabilities, including calculating probabilities and expected counts within a contingency table, along with a chi-square test. The final question explores a regression model, interpreting the coefficients, constructing confidence intervals for the slope and intercept, and calculating t-statistics and p-values. The assignment demonstrates the application of statistical methods to real-world scenarios, including market research and data analysis.

STATISTICS
[Document subtitle]
[DATE]
[Company name]
[Company address]

Question 1
1) 99% confidence interval for p
Sample size = 150
Number of people found disappointed = 35
Proportion = 35/150 =0.233
The z value for 99% confidence interval = 2.5758
Lower limit of 99% confidence interval
Upper limit of 99% confidence interval
99% confidence interval = [0.14438 0.32228]
It can be said with 99% confidence that the sample proportion of people found disappointed
with new Tesla would fall between 0.14438 and 0.32228.
2) Null and alternative hypothesis
Sample proportion of people whose expectation were met p* = (150-35)/150 = 0.76667
The test statistics z is computed as shown below.
The test statistics z comes out 2.99572.
1

3) It is a right tailed p value and hence, the p value
The p value = 0.001369
4) Significance level = 0.05
Here, p value is lower than significance level and hence, the null hypothesis will be rejected.
“False”
5) 95% confidence interval
Sample size of Torontonian = 100
Number of Torontonian planned to purchase electric car = 18
P1 = 18/100=0.18
Sample size of Montrealers = 100
Number of Montrealers planned to purchase electric car = 16
P2 = 16/100=0.16
The z value for 95% confidence interval = 2.5758
Now,
Lower limit of 95% confidence interval
Upper limit of 95% confidence interval
95% confidence interval = [0.096 0.304]
It can be said with 95% confidence that the difference of proportion of Torontonian and
Montrealers would fall between 0.14438 and 0.32228.
2

Question 2
1) The value c would denote the number of columns whose value would be 3.
2) Requisite probability = (73/350) = 0.209
3) Requisite probability = (77/150) = 0.513
4) The expected count of Quebec-born Canadians that prefer Comedy = (120*117)/350 =
40.11
5) Expected count of cell (Quebec and Drama)
Expected count= 120∗160
350 =54.857∨ 384
7
Chi square value= ( 59− 384
7 )
2
384
7
=0.3129
6) Number of degree of freedom = (r-1)*(c-1) = (3-1)*(3-1) = 4
7) The chi square statistics and p value of test
Chi square statistics = 10.79
3

p value of test = 0.0290
Question 3
1) The annual bonus would increase by $10,438.76 if the work experience tends to increase
by one year. This is evident since this is the value of the slope coefficient of the
regression model whose output has been presented.
2) The 95% confidence interval for the slope is found as shown below.
Relevant t value considering df = 50-1= 49 and significance level of 5% is 2.0096
Standard error for slope coefficient = 1.178
Margin of error = 1.178*2.0096 = 2.367
Lower limit of 95% confidence interval for slope coefficient = Mean – Margin of error =
10.439 – 2.367 = 8.072
Upper limit of 95% confidence interval for slope coefficient = Mean + Margin of error =
10.439 + 2.367 = 12.806
3) The 95% confidence interval for the intercept is found as shown below.
Relevant t value considering df = 50-1= 49 and significance level of 5% is 2.0096
Standard error for intercept = 2.803
Margin of error = 2.803*2.0096 = 5.633
Lower limit of 95% confidence interval for intercept = Mean – Margin of error = -8.64 -5.633
= -14.27
Upper limit of 95% confidence interval for intercept = Mean + Margin of error = -8.64 +
5.633 = -3.01
4) The intercept value is -8.64 which indicates the value of the annual bonus when the
experience is zero. Clearly, in the given case the intercept does not have practical utility
as annual bonus cannot be negative.
4

5) Value of A = Regression slope coefficient/Standard error of slope coefficient
Hence, value of A = (10.438763/1.1782) = 8.86
6) The t statistics for the intercept is -3.08256. The degree of freedom is (50-1) = 49.
Considering the above information provided and using the t table, the p value is 0.0034.
Hence, value of B is 0.0034.
5