BCOR 1020Business Statistics
Lecture 20 – April 3, 2008
Overview
• Chapter 9 – Hypothesis Testing– Problem: Testing A Hypothesis on a
Proportion– Testing a Mean (): Population Variance ()
Known– Problem: Testing a Mean () when Population
Variance () is Known
Problem: Testing A Hypothesis on a Proportion
Suppose we revisited our earlier example and conducted an additional survey to determine whether the proportion of our target market that is willing to pay $25 per unit exceeds 20% (as required by the business case).• In our new survey, 72 out of 300 respondents said they
would be willing to pay $25 per unit.• At the 5% level of significance, conduct the appropriate
hypothesis test to determine whether the population proportion exceeds 20%.
• Include the following:– State the level of significance, .– State the null and alternative hypotheses, H0 and H1.– Compute the test statistic– State the decision criteria– State your decision
(Overhead)
Problem: Testing A Hypothesis on a Proportion
Work: =
H0:
H1:
Test Statistic (and Distribution under H0)…
Decision Criteria…
Decision…
Clickers
What are the appropriate null and alternative hypotheses?
(B) H0: >
H1: <
(A) H0: <
H1: >
(C) H0: =
H1:
Clickers
What is the point estimate of the population proportion ?
(A) p = 0.15
(B) p = 0.20
(C) p = 0.22
(D) p = 0.24
(E) p = 0.36
Clickers
What is the calculated value of your test statistic?
(A) Z* = 2.00
(B) Z* = 1.73
(C) Z* = 0.87
(D) Z* = 0.71
Clickers
What is your decision criteria?
(A) Reject H0 if Z* < -1.645
(B) Reject H0 if Z* < -1.960
(C) Reject H0 if Z* > 1.960
(D) Reject H0 if Z* > 1.645
(E) Reject H0 if |Z*| > 1.960
Problem: Testing A Hypothesis on a Proportion
Conclusion:• Since our test statistic Z* = 1.73 > Z = 1.645,
we will reject H0 in favor of H1: > 0.20.
• Based on the data in this sample, there is statistically significant evidence that the population proportion exceeds 20%.
Chapter 9 – Testing a Mean ( known)
Hypothesis Tests on ( known):• If we wish to test a hypothesis about the mean of a
population when is assumed to be known, we will follow the same logic…
– Specify the level of significance, (given in problem or assume 10%).
– State the null and alternative hypotheses, H0 and H1 (based on the problem statement).
– Compute the test statistic and determine its distribution under H0.
– State the decision criteria (based on the hypotheses and distribution of the test statistic under H0).
– State your decision.
Chapter 9 – Testing a Mean ( known)
Selection of H0 and H1:• Remember, the conclusion we wish to test should be
stated in the alternative hypothesis.• Based on the problem statement, we choose from…
(i) H0: > 0
H1: < 0
(ii) H0: < 0
H1: > 0
(iii) H0: = 0
H1: 0
where 0 is the null hypothesized value of (based on the problem statement).
Chapter 9 – Testing a Mean ( known)
Test Statistic:• Start with the point estimate of , in xx 1
• Recall that for a large enough n {n > 30}, is approximately normal with x n
x
xand
• If H0 is true, then is approximately normal with
0 x nx
and
x
• So, the following statistic will have approximately a standard normal distribution:
n
xZ
0*
Chapter 9 – Testing a Mean ( known)
Decision Criteria:• Just as with tests on proportions, our decision criteria will
consist of comparing our test statistic to an appropriate critical point in the standard normal distribution (the distribution of Z* under H0).
(i) For the hypothesis test H0: > 0 vs. H1: < 0, we will reject H0 in favor of H1 if Z* < – Z.
(ii) For the hypothesis test H0: < 0 vs. H1: > 0, we will reject H0 in favor of H1 if Z* > Z.
(iii) For the hypothesis test H0: = 0 vs. H1: 0, we will reject H0 in favor of H1 if |Z*| > Z/2.
Chapter 9 – Testing a Mean ( known)
Example (original motivating example):• Suppose your business is planning on bringing a
new product to market.• There is a business case to proceed only if
– the cost of production is less than $10 per unit
and– At least 20% of your target market is willing to pay $25
per unit to purchase this product.
• How do you determine whether or not to proceed?
Chapter 9 – Testing a Mean ( known)
Motivating Example (continued) :• Assume the cost of production can be modeled
as a continuous variable.– You can conduct a random sample of the
manufacturing process and collect cost data.– If 40 randomly selected production runs yield and
average cost of $9.00 with a standard deviation of $1.00, what can you conclude?
– We will test an appropriate hypothesis to determine whether the average cost of production is less than $10.00 per unit (as required by the business case).
(Overhead)
Chapter 9 – Testing a Mean ( known)
Motivating Example (continued) :• Conduct the Hypothesis Test:
– Specify , say = 0.05 for example.– Select appropriate hypotheses.
• Since we want to test that the mean is less than $10, we know that 0 = 10 and H1 should be the “<“ inequality.
• So we will test (i) H0: > 0 vs. H1: < 0.
– Calculate the test statistic…
401
0* 109
n
xZ
32.6* Z
Chapter 9 – Testing a Mean ( known)
Motivating Example (continued) :• Conduct the Hypothesis Test:
– Use the Decision criteria:
• (i) we will reject H0 in favor of H1 if Z* < – Z, where Z.05 = 1.645.
– State our Decision…
• Since Z* = -6.32 < -Z.05 = -1.645, we reject H0 in favor of H1.
– In “plain” language…• There is statistically significant evidence that the
average cost of production is less than $10 per unit.
Chapter 9 – Testing a Mean ( known)Calculating the p-value of the test:• The p-value of the test is the exact probability of
a type I error based on the data collected for the test. It is a measure of the plausibility of H0.– P-value = P(Reject H0 | H0 is True) based on our data.– Formula depends on which pair of hypotheses we are testing…
*2 ZZPvaluep
(i) For the hypothesis test H0: > 0 vs. H1: < 0,
(ii) For the hypothesis test H0: < 0 vs. H1: > 0,
(iii) For the hypothesis test H0: = 0 vs. H1: 0,
)( *ZZPvaluep
)( *ZZPvaluep
Chapter 9 – Testing a Mean ( known)
• Example: Let’s calculate the p-value of the • test in our example…• We found Z* = –6.32
• Since we were testing H0: > 0 vs. H1: < 0,
0)32.6()( * ZPZZPvaluep
Interpretation:If we were to reject H0 based on the observed data, there is approximately zero probability that we would be making a type I error. Since this is smaller than = 5%, we will reject H0.
Problem: Testing A Hypothesis on a Mean ( Known)
In an effort to get a loan, a clothing retailer has made the claim that the average daily sales at her store exceeds $7500. Historical data suggests that the purchase amount is normally distributed with a standard deviation of = $1500.• In a randomly selected sample of 24 days sales data, the
average daily sales were found to be = $7900.• At the 10% level of significance, conduct the appropriate
hypothesis test to determine whether the data supports the retailer’s claim.
• Include the following:– State the level of significance, .– State the null and alternative hypotheses, H0 and H1.– Compute the test statistic– State the decision criteria– State your decision
x
(Overhead)
Problem: Testing A Hypothesis on a Mean ( Known)
Work: =
H0:H1:
Test Statistic (and Distribution under H0)…
Decision Criteria…
Decision…
Clickers
What are the appropriate null and alternative hypotheses?
(A) H0: >
H1: <
(B) H0: <
H1: >
(C) H0: =
H1:
Clickers
What is the calculated value of your test statistic?
(A) Z* = 0.27
(B) Z* = 1.28
(C) Z* = 1.31
(D) Z* = 1.96
(E) Z* = 5.33
Clickers
What is your decision criteria?
(A) Reject H0 if Z* < -1.645
(B) Reject H0 if Z* < -1.282
(C) Reject H0 if Z* > 1.282
(D) Reject H0 if Z* > 1.645
(E) Reject H0 if |Z*| > 1.960
Clickers
What is your decision?
(A) Reject H0 in favor of H1
(B) Fail to reject (Accept) H0 in favor of H1
(C) There is not enough information
Problem: Testing A Hypothesis on a Mean ( Known)
Conclusion:• Since our test statistic Z* = 1.31 is greater than
Z = 1.282, we will reject H0 in favor of H1: > $7500.
• Based on the data in this sample, there is statistically significant evidence that the average daily sales at her store exceeds $7500.
Clickers
Given our test statistics Z* = 1.31, calculate the p-value for the hypothesis test H0: < 7500 vs. H1: > 7500.
(A) p-value = 0.0060
(B) p-value = 0.0951
(C) p-value = 0.9940
(D) p-value = 0.9049