Assignment definition and meaning | Assignment

Michael Langdon
20003727
Name:
161.101 Statistics for Business
ID Number:
ASSIGNMENT 3 DUE 3 FEBRUARY 2020
Q1 Consider again the situation described in Assignment 2 Question 5: from a random
sample of 60 motorists during a particular week, 9 had purchased diesel.
The owner wants to test whether this gives evidence to contradict his belief that 10%
of customers use diesel.
(a) State, in words and in symbols, the hypotheses to be tested. [2]
(b) What is the p-value? [1]
(c) Is there evidence to reject the null hypothesis? Justify your answer. [1]
(d) State the conclusion in context. [2]
(e) Compare this analysis to that of Assignment 2 Question 5(c). Discuss. [2]
161.101/Ass3/1903 1
Ho : p = 10%
Ha : p ≠ 10%
n = 60
significance = 0.05
pbar = sample proportion = 9/60 = 0.15
Z = (pbar – p0)/sqrt((p0(1-p0)/n) = (0.15 – 0.1)/sqrt((0.1(1-0.1)/60) = 1.29
P value for Z = 1.29 = twice of area of normal curve in right of Z=1.29 = 0.197051
As the p value is over chosen significance level of 0.05, hence, there is not sufficient
evidence to reject the null hypothesis.
Hence, there is not sufficient evidence to conclude that the proportion of customers
who use diesel is not equal to 10%.
In assignment 2 part c also similar results found where there was not sufficient
evidence to reject the null hypothesis.

Michael Langdon
20003727
Name:
161.101 Statistics for Business
ID Number:
Q2 Consider again the data file NZ exports 2013 and 2014.xlsx. Suppose we want to
carry out a formal hypothesis test of whether the value of NZ exports to these
countries has increased from 2013 to 2014.
(a) State, in words and in symbols, the hypotheses being tested. [3]
(b) State the appropriate p-value. [1]
(c) Is there evidence to reject the null hypothesis? Justify your answer. [1]
(d) State the conclusion in context. [2]
(e) Examine an appropriate normal probability plots. What does it suggest? [3]
161.101/Ass3/1903 2
μ1 = population mean of 2013 exports
μ2 = population mean of 2014 exports
H0: μ2 – μ1 <= 0
H1: μ2 – μ1 > 0
Z = ((x2bar – x1bar) – (μ2 – μ1))/σ, σ = sqrt(σ1^2/n1 + σ2^2/n2)
Now, p value = 0.001017
As the p value is less than 0.05 thus there is enough evidence to reject the null
hypothesis.
Now, as there is enough evidence to reject the null hence, value of NZ exports to
given countries has increased from 2013 to 2014.
The normal probability plot of the difference of 2013 and 2014 export data is created in excel as given below.
From the normal probability plot it can be seen that the points are not aligned in the normal line and thus the
sample does not belong from a population which is normally distributed.

Michael Langdon
20003727
Name:
161.101 Statistics for Business
ID Number:
Q3 Consider again the situation described in Assignment 2 Question 1. Suppose we want
to carry out a formal test of whether Loyalty is related to Length of Service.
(a) State the hypotheses to be tested. [2]
(b) Calculate the expected frequencies. [1]
(c) What is the p-value? [1]
(d) State the conclusion in context, justifying your answer. [2]
(e) Calculate the proportion who would leave for each age group. What does it
suggest? [2]
161.101/Ass3/1903 3
H0: Loyalty and length of service are uncorrelated
H1: There exist significant correlation between loyalty and length of service.
Length of Service
Loyalt
y < 1 1 - 5 6 - 10 > 10
Remai
n 3 5 7 9
Leave 9 7 5 3
P value is 0.00125.
Hence, there is significant evidence to reject the null hypothesis and it can be
concluded that Loyalty and length of service are related.
The proportion which leaves for less than 1 years of length of service age is 9/12 =
75%, the proportion who leaves in length of service between 1 and 5 years is 7/12 =
58.33%, the proportion that leaves for age group 6-10 is 41.67% and proportion that
leaves in length of service > 10 is 25%. This indicates that as length of service
increases the number of people leaves become less or more number of people remain.