Docsity
Docsity

Prepare for your exams
Prepare for your exams

Study with the several resources on Docsity


Earn points to download
Earn points to download

Earn points by helping other students or get them with a premium plan


Guidelines and tips
Guidelines and tips

Hypothesis Testing on Canadian Non-Mortgage Debt and Apple Care Wait Times, Assignments of Statistics

Three statistical hypothesis testing problems related to the average non-mortgage debt of canadians, the average wait time for the apple care support line, and the proportion of adults participating in fitness activities at least twice a week. Details on the problem context, the null and alternative hypotheses, the calculation of test statistics and p-values, and the interpretation of the results to draw conclusions. The topics covered in this document are relevant to university-level courses in statistics, data analysis, and business analytics. The document could be useful for university students as study notes, lecture notes, or exam preparation materials, particularly for courses in statistics, data analysis, and business analytics.

Typology: Assignments

2023/2024

Uploaded on 04/15/2024

jhjk-lkhjkh
jhjk-lkhjkh 🇨🇦

1 document

1 / 4

Toggle sidebar

This page cannot be seen from the preview

Don't miss anything!

bg1
STAT 1181 Assignment 5 (40 marks) Due April 4, 2024
1. According to research, the average non-mortgage debt of Canadians is $23000. A simple random
sample of 38 Canadians was selected, and their average non-mortgage debt was found to be
$21000. Assume the population standard deviation of non-mortgage debt is $4600. At the 0.01
level of significance, is there sufficient evidence to conclude that the average non-mortgage debt
of Canadians has decreased? (10 marks)
a) Define the parameter of interest in the context of the question. (1 mark)
Ans-: The parameter of interest is mu ,the true mean non-mortgage debt of all Canadians.
b) Set up the null and alternative hypotheses in terms of the parameter. (2 marks)
Sample mean=$21000
Population mean under the null hypothesis=$23000
Population standard deviation=$4600
Sample size=38
Z= (21000-23000)/(4600/ √38)
Z value = -2.68
H0 = mu=$23000 The average non-mortgage debt among Canadians is 23000, indicating no
change.
Ha = mu<$23000 The average non-mortgage debt among Canadians has decreased from 23000.
c) Calculate the test statistic. Find the p-value. Check the required condition(s) to validate the
pvalue. (3 marks)
Ans
Looking at the distribution table lower tail P value is approximately 0.0037.
d) Interpret the p-value in the context of the question. (2 marks)
0.0037 indicates a very low probability of observing a sample mean
These rejects the null hypothesis and conclude that the average non-mortgage debt of Canadians
has decreased from 23000. This suggest strong evidence against the null hypothesis at 0.01 level
of significance.
e) What conclusion can be drawn? (2 marks)
Answer
Since the p-value is less than the significance level of 0.01, we reject the null hypothesis.
Therefore, there is sufficient evidence to conclude that the average non-mortgage debt of
Canadians has decreased from the previously assumed average of $23,000.
2. Apple claims that the average wait time for a customer calling the Apple Care support line is 180
seconds. A simple random sample of 35 customers was selected, and their average wait time was
found to be 190 seconds. Assume the population standard deviation of wait time is 45 seconds.
At the 5% level of significance, is there enough evidence to conclude that the average wait time
for a customer calling the Apple Care support line is more than 180 seconds? (10 marks)
a) Define the parameter of interest in the context of the question. (1 mark)
The parameter of interest is mu, the true mean wait time for all customers calling the Apple Care
support line.
pf3
pf4

Partial preview of the text

Download Hypothesis Testing on Canadian Non-Mortgage Debt and Apple Care Wait Times and more Assignments Statistics in PDF only on Docsity!

STAT 1181 Assignment 5 (40 marks) Due April 4, 2024

  1. According to research, the average non-mortgage debt of Canadians is $23000. A simple random sample of 38 Canadians was selected, and their average non-mortgage debt was found to be $21000. Assume the population standard deviation of non-mortgage debt is $4600. At the 0. level of significance, is there sufficient evidence to conclude that the average non-mortgage debt of Canadians has decreased? (10 marks) a) Define the parameter of interest in the context of the question. (1 mark) Ans-: The parameter of interest is mu ,the true mean non-mortgage debt of all Canadians. b) Set up the null and alternative hypotheses in terms of the parameter. (2 marks) Sample mean=$ Population mean under the null hypothesis=$ Population standard deviation=$ Sample size= Z= (21000-23000)/(4600/ √38) Z value = - 2. H 0 = mu=$23000 The average non-mortgage debt among Canadians is 23000, indicating no change. Ha = mu<$23000 The average non-mortgage debt among Canadians has decreased from 23000. c) Calculate the test statistic. Find the p - value. Check the required condition(s) to validate the p value. (3 marks) Ans Looking at the distribution table lower tail P value is approximately 0.0037. d) Interpret the p - value in the context of the question. (2 marks) 0.0037 indicates a very low probability of observing a sample mean These rejects the null hypothesis and conclude that the average non-mortgage debt of Canadians has decreased from 23000. This suggest strong evidence against the null hypothesis at 0.01 level of significance. e) What conclusion can be drawn? (2 marks) Answer Since the p-value is less than the significance level of 0.01, we reject the null hypothesis. Therefore, there is sufficient evidence to conclude that the average non-mortgage debt of Canadians has decreased from the previously assumed average of $23,000.
  2. Apple claims that the average wait time for a customer calling the Apple Care support line is 180 seconds. A simple random sample of 35 customers was selected, and their average wait time was found to be 190 seconds. Assume the population standard deviation of wait time is 45 seconds. At the 5% level of significance, is there enough evidence to conclude that the average wait time for a customer calling the Apple Care support line is more than 180 seconds? (10 marks) a) Define the parameter of interest in the context of the question. (1 mark) The parameter of interest is mu, the true mean wait time for all customers calling the Apple Care support line.

b) Set up the null and alternative hypotheses in terms of the parameter. (2 marks) H 0 : mu=180 seconds. This states that the average wait time for the Apple Care support line is 180 seconds, aligning with Apple's claim. Ha: mu>180 seconds. This suggest that the average wait time for the Apple Care support line is more than 180 seconds, contrary to Apple ‘s claim. c) Calculate the test statistic. Find the p - value. Check the required condition(s) to validate the p value. (3 marks) Answer sample mean = 190 seconds population mean under the null hypothesis = 180 seconds population standard deviation = 45 seconds sample size = 3 5 z = (190-180)/(45/ √35) z value approximately 1. The p value is the 0.0943 by looking at the upper tail of the normal distribution. d) Interpret the p - value in the context of the question. (2 marks) The p value of approximately 0.0943 indicates the probability of observing the p value is greater than the alpha level 0.05, there enough evidence to reject the null hypothesis at the 5% level of significance. e) What conclusion can be drawn? (2 marks) The p-value 0.0943 is greater than the significance level of 0.05, fail to reject the null hypothesis. There is not sufficient evidence at the 5% level of significance to conclude that the average wait time for a customer calling the Apple Care support line is more than 180 seconds.

  1. According to research, 35% of adults participate in fitness activities at least twice a week. However, these fitness activities change as people get older, and occasionally participants become non-participants as they age. In a recent survey of 100 randomly selected adults, 30 indicated that they participated in a fitness activity at least twice a week. At the 0.1 significance level, can we conclude that less than 35% of adults participate in fitness activities at least twice a week? (10 marks) a) Define the parameter of interest in the context of the question. (1 mark) The parameter of interest is p, the true proportion of adults participate in fitness activities at least twice a week. b) Set up the null and alternative hypotheses in terms of the parameter. (2 marks) H 0 : p=0.35.This states that 35% of adults participate in fitness activities at least twice a week, aligning with the initial research findings. Ha: p<0.35.This suggest that less than 35% of adults participate in fitness activities at least twice a week, indicating a potential decline as people age. c) Calculate the test statistic. Find the p - value. Check the required condition(s) to validate the p value. (3 marks) Sample proportion (30/100 = 0.3) Population proportion under the null hypothesis = 0.35. Sample size = 100 Z= (0.3-0.35)/ √(0.35(1-0.35/100)