Assignment 5 Assignment 5 consists of a theory component (5A, worth 80%) and a computer component (5B, worth 20%). The grades allocated are summarized below. Assignment Grade weight (%) Assignment 5A....

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Answer To: Assignment 5 Assignment 5 consists of a theory component (5A, worth 80%) and a computer component...

Mohd answered on Oct 07 2021
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Assignment 5
Assignment 5 consists of a theory component (5A, worth 80%) and a computer component (5B, worth 20%).
The grades allocated are summarized below.
    Assignment
    Grade weight (%)
    Assignment 5A. Theory
    80
    Assignment 5B. Computer
    20
    Total marks
    100
You must submit Assignment 5A and Assignment 5B together as PDFs, through the appropriate drop box on the course home page. Submit:
One PDF solution file (file) entitled Assignment5A containing all your answers to Assignment 5A, presented in the proper order. Your name and student ID number must be at the top of the first page of your solution file.
and
One PDF solution file (file) entitled As
signment5B containing all your answers to Assignment 5B, presented in the proper order.
We suggest that you print the Assignment 5 questions, so that you can conveniently review the questions with their solutions when you prepare for your exams.
Assignment 5A. Theory Component
Show your work for this component. Where relevant andunless otherwise instructed, keep your calculations and your final answer to at least four decimals.For any test of hypothesis questions, show all steps in detail and provide a conclusion in terms of the context of the question.
Problem 1. Salaries(23 marks)
A study was conducted to determine if the salaries of the professors from two neighbouring universities were equal. A sample of 20 professors from each university was randomly selected. The mean from the first university was $109,100 with a population standard deviation of $2300. The mean from the second university was $110,500 with a population standard deviation of $2100.Assume that the distribution of professor salaries, at both universities, are approximately normally distributed. Level of significance = 0.05.
    University
    Sample Size
    Mean ( in $)
    SD (in $)
    A
    20
    109100
    2300
    B
    20
    110500
    2100
Using the critical value approach test the claim that the salaries from both universities are equal. Include all key steps of the test. [10 marks]
Step 1: Specify the null and alternate hypothesis
H0: μ1= μ2 Salaries of the professors from two neighbouring universities were equal.
HA: μ1≠ μ2 Salaries of the professors from two neighbouring universities were not equal.
Step 2: Significance level of the test: 0.05
Step 3:
Using the sample data and assuming the null hypothesis is true, calculate the value of the test statistic.
t=(Xa−Xb)/( √(sa2/na+ sb2/nb))
Which follows a t-distribution with n - 1 degrees of freedom.
test statistic = 2.0102
Significance level of the test: 0.05 Critical value(-t0.025,19) = 2.0930
Step4: Compare and interpret critical value and test statistics
There are two critical values for the two-tailed test H0 : μ1= μ2 versus HA : μ1≠ μ2 one for the left-tail denoted -t(α/2, n - 1) and one for the right-tail denoted t(α/2, n - 1). The value -t(α/2, n - 1) is the t-value such that the probability to the left of it is α/2, and the value t(α/2, n - 1) is the t-value such that the probability to the right of it is α/2.
It can be shown using either statistical software or a t-table that the critical value -t0.025,19 is 2.0930 and the critical value t0.025,19 is 2.0930. That is, we would accept the null hypothesis H0 : μ1= μ2 the test statistic t* is less than 2.0930.
Use the four-step P-value approach and test the claim that the salaries from both universities are equal. [10 marks]
Step 1: Specify the null and alternate hypothesis
H0: μ1= μ2 salaries of the professors from two neighbouring universities were equal.
HA : μ1≠ μ2 salaries of the professors from two neighbouring universities were not equal.
Step 2: Significance level of the test: 0.05
Step 3:
Using the sample data and assuming the null hypothesis is true, calculate the value of the test statistic.
t=(Xa−Xb)/( √(sa2/na+ sb2/nb))
which follows a t-distribution with n - 1 degrees of freedom.
test statistic = 2.0102
Significance level of the test: 0.05 p[z]=0.9778
Pvalue=1-0.9778=0.022
p-value is less than 0.025 or 0.05. we accept the null hypothesis.
Are the conditions met for using this test? [3 marks
Yes, following conditions are met.
· Random sample
· Independence
· Normal distribution
· Parent population normal
Problem 2. Flexibility (12 marks)
Are women more flexible than men? To answer this question, a physical therapist measured the flexibility of 15 randomly selected women and 20 randomly selected men by determining the number of inches subjects could reach while sitting on the floor with their legs straight out and back perpendicular to the ground. The more flexible an individual is, the more they can reach from the seated position. The physical therapist obtained the following data:
    Gender
    Sample Size
    Mean ( in inches)
    SD (in inches)
    Women
    15
    20.86
    2.11
    Men
    20
    18.43
    3.16
Assume that both population distributions are approximately normal and that the population variances are NOT equal.
1. Is the sampling dependent or independent in this study? Justify your answer. [2 marks]
Its independent. In this sample we have no specified relationship. Fully isolated variable with zero influence.
Test the hypothesis at α = 0.05. Using the critical value approach, include all key steps of test of hypothesis.[10 marks]
Given x1 = 20.86, x2 = 18.43, n1 = 15, n2 = 20, s 1 = 2.11, s 2 = 3.16 and the hypothesis of interest is
H0 : m1 = m2
    H A : m1 > m2
Under H0 , the test statistic
Z=(Xa−Xb)/( √(sa2/na+ sb2/nb))
Z=(20.86−18.43)/( √(2.112/15+ 3.162/20))
Z=2.723513
Since Z = 2.72513 > Za = 1.645 at a = 0.05 level, H0 is to be rejected, i.e., we conclude that women are more flexible than men.
Problem 3. Customer waiting times (10 marks)
Bank A claims that the waiting time for its customers is less than its competitor Bank B’s customers. A researcher collected data on the wait time for customers from both banks.Test the claim. Assume the samples are random and independent, and the populations are normally distributed. Also, assume that the unknown population variances are equal.Test the claim at 0.01 level of significance.
    Bank A
    Bank B
    = 15
= 5.3 minutes
= 1.1 minutes
     = 16
= 5.6 minutes
= 1.0 minutes
Is there sufficient evidence to believe that the wait time for Bank A is less than for Bank B?
Include all key steps of test of hypothesis, using the critical value approach. [10 marks]
We want to test
H0 : m1 = m2 vs HA : m1 < m 2 .
Under H0 , the test statistic~ (0,1)
Z=(Xa−Xb)/( √(sa2/na+ sb2/nb))
Z=(5.3−5.6)/( √(1.12/15+ 1.02/16))
Since Z = -0.7928 < Za = -2.33 at a = 0.01 level, H0 is to be rejected and we conclude
that on wait time for Bank A is equal to Bank B.
Problem 4....
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