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Quantitative Methods

QMS 202

Jason Chin- Tiong Chan

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.

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