Class Notes (834,935)
QMS 202 (64)
Lecture

# Chapter 10

18 Pages
176 Views

School
Department
Quantitative Methods
Course
QMS 202
Professor
Jason Chin- Tiong Chan
Semester
Winter

Description
Business Statistics II Chapter10 Chapter10 Fundamentals of Hypothesis Testing: One-Sample tests Outcomes: 1. Define a hypothesis and hypothesis testing 2. Describe the hypothesis-testing procedure 3. Distinguish between a one-tailed and a two-tailed test of hypothesis 4. Conduct a hypothesis test for a population mean with known population standard deviation 5. Conduct a hypothesis test for a population mean with unknown population standard deviation 6. Conduct a hypothesis test for a population proportion 7. Define Type I and Type II errors 8. Compute and interpret a p-value What is a Hypothesis? 1. A hypothesis is a statement about a population. 2. In statistical analysis we make a claim, that is, state a hypothesis, collect data, and then use the data to test the claim. 3. In most cases the population is so large that it is not feasible to study all the items, objects or persons in the population. For example, it would not be possible to contact every accountant in Canada to find out his or her annual income. 4. An alternative to measuring or interviewing the entire population is to take a sample from the population and test a statement to determine whether the sample does or does not support the statement concerning the population. What is Hypothesis Testing? 1. A procedure based on sample evidence and probability theory which determine whether the hypothesis is a reasonable statement. 2. The procedure for testing a hypothesis Step1. Define the parameter(s) of interest Step2. State null and alternative hypotheses Step3. Identify a level of significance Winter2011 page#1 www.notesolution.com Business Statistics II Chapter10 Step4. Identify the test statistic Step5. Compute the value of test statistic and p-value Step6. State the statistical decision and business conclusion Note: 1. Null Hypothesis H o : A statement about the value of a population parameter Alternative Hypothesis H A: A statement that is accepted if the sample data provide evidence that the null hypothesis is false 2. Level of significance: The probability of rejecting the null hypothesis when it is true 3. Test Statistics: A value, determined from sample information, used to determine whether or not to reject the null hypothesis 4. p-value : the probability of observing a sample value is as extreme as, or more extreme than, the value observed, when the null hypothesis is true 5. Type I error: Rejecting the null hypothesis when it is true 6. Type II error: Do not reject the null hypothesis when it is false Winter2011 page#2 www.notesolution.com Business Statistics II Chapter10 Example1 The BigMacBurger restaurant chain claims that the waiting time of all customers for service is normally distributed, with a population mean of waiting time of 4 minutes and a population standard deviation of 1.4 minutes. The quality assurance department found in a sample of 60 customers at Yonge and Dundas that the mean waiting time was 4.1 minutes. Can we conclude that the mean waiting time is more than 4 minutes? Test at 5% of level of significance. Calculator Output 1-Sample z Test μ > 4 z = 0.55328 p = 0.29003 x = 4.1 n = 60 Step1 Let μ be the population mean of waiting time Step2 H o : μ ≤ 4 H A: μ > 4 Step3 Level of significance = 0.05 Step 4 one-sample mean z test Step5 z stat 0.55328 p-value = 0.29003 Winter2011 page#3 www.notesolution.com Business Statistics II Chapter10 x − μ 4 .1 − 4 zstat= σ = 1.4 = 0.55328 p-value = P(z > 0.55328 = ) n 60 0.29003 zcriti cal1.64485 Step6 Since the p-value > 0.05, do not reject the null hypothesis. There is not enough evidence to conclude that the population mean of waiting time is more than 4 minutes. Example2 A recent survey found that young professional working on Bay Street watched an average of 6.8 DVDs per month. A random sample of 36 young professionals revelaed that the mean number of DVDs watched last month was 6.2, with a population standard deviation of 2.5. Can we conclude that young professionals on Bay Street watch fewer than 6.8 DVDs? Use α =0.05 Calculator Output 1-Sample z Test μ < 6.8 z = -1.44 p = 0.074933 x = 6.2 n = 36 Step1 Let μ Step2 H o : μ H A: μ Step3 Level of significance = Step 4 Step5 Winter2011 page#4 www.notesolution.com Business Statistics II Chapter10 zstat= p-value = x − μ z = σ = stat n zcriti cal Step6 T Test of Hypothesis for the Mean (population standard deviation σ unknown) Example3 The Ryerson’s Discount Appliance Store issues its own credit card. The credit card manager wants to find whether the mean monthly-unpaid balance is more than \$500. The level of significance is set at 0.05. A random check of 180 unpaid balances revealed the sample mean is \$508 and the standard deviation of the sample is \$47. It is known that the monthly-unpaid balance is normally distributed. Should the credit card manager conclude that the population mean is greater than \$500? Calculator Output 1-Sample t Test μ > 500 t = 2.2836 p = 0.011783 x = 508 sx = 47 n = 180 Step1 Let μ Step2 H o : μ H : μ A Step3 Level of significance = Winter2011 page#5 www.notesolution.com Business Statistics II Chapter10 Step 4 Step5 tstat= p-value = x − μ t stat s = n d . f = tcriti cal Step6 Example4 At the time John was hired as a server at the Ryerson Family restaurant at Dundas and Bay, he was told” You can earn average of not more than \$120 a day in tips.” Over the first 40 days he was employed at the restaurant, the mean daily amount of his tips was \$120.50 and a standard deviation of \$41.32. It is known that the daily amount of tips is normally distributed. At the 0.05 significance level, can John conclude that he is earning an average of more than \$120 in tips per day in the entire year? Calculator Output 1-Sample t Test μ > 120 t = 0.076531 p = 0.46969 x = 120.5 sx = 41.32 n = 40 Step1 Let μ Step2 H : μ o H A : μ Step3 Level of significance = Winter2011 page#6 www.notesolution.com Business Statistics II Chapter10 Step 4 Step5 t stat= p-value = x − μ t stat= s = n d . f = tcriti cal Step6 Example 5 A national grocer’s magazine reports that the typical shopper spends nine minutes in line waiting to check out. A sample of 30 shoppers at the local Yes Frills showed a mean of 9.1 minutes with a standard deviation of 2.9 minutes. It is known that the waiting time at local Yes Frills is normally distributed. Is the waiting time at the local Yes Frills different from that reported in the national magazine? Use the 0.05 significance level.
More Less

Related notes for QMS 202
Me

OR

Join OneClass

Access over 10 million pages of study
documents for 1.3 million courses.

Join to view

OR

By registering, I agree to the Terms and Privacy Policies
Just a few more details

So we can recommend you notes for your school.