Transcript Slides 2-5 Hypothesis Testing

BA 275
Quantitative Business Methods
Agenda
 Quiz #3
 Statistical Inference: Hypothesis Testing
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Types of a Test
P-value
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Central Limit Theorem (CLT)
 The CLT applied to Means
If X ~ N (  ,  2 ) , then X ~ N (  ,
2
).
n
If X ~ any distribution with a mean , and variance 2,
then X ~ N (  ,
2
n
) given that n is large.
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Example 1
 How much time do executives spend each
day reading and sending e-mail? A survey of
162 executives was conducted and the mean
time (in minutes) was 63.6975 minutes.
 Assume that the population std is 18.9403.
Can we infer that the mean amount of time
spent by all executives reading and sending
e-mail exceeds 60 minutes?
 Assume 5% significance level.
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Example 2
 How much time do executives spend each
day reading and sending e-mail? A survey of
162 executives was conducted and the mean
time (in minutes) was 63.6975 minutes.
 Assume that the population std is 18.9403.
Can we infer that the mean amount of time
spent by all executives reading and sending
e-mail is different from 60 minutes?
 Assume 5% significance level.
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Example 3
 How much time do executives spend each
day reading and sending e-mail? A survey of
162 executives was conducted and the mean
time (in minutes) was 63.6975 minutes with a
standard deviation of 18.9403.
 At 5% significance level, we concluded that
the mean amount of time exceeds 60
minutes.
 By how much?
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The p-Value Approach
 (textbook, p.386) The p-value is the
probability, under the assumption that H0 is
true, of obtaining a test statistic as or more
extreme than the one actually obtained from
the data.
 (alternative definition) The p-value is the
smallest value of a that would lead to the
rejection of H0.
 The smaller the p-value, the stronger the
evidence against H0 provided by the data.

Compare the p-value to the significance level a.
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Example 1 (cont’d)
 How much time do executives spend each
day reading and sending e-mail? A survey of
162 executives was conducted and the mean
time (in minutes) was 63.6975 minutes.
 Assume that the population std is 18.9403.
Can we infer that the mean amount of time
spent by all executives reading and sending
e-mail exceeds 60 minutes?
 Assume 5% significance level.
 Calculate the p-value.
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Example 2 (cont’d)
 How much time do executives spend each
day reading and sending e-mail? A survey of
162 executives was conducted and the mean
time (in minutes) was 63.6975 minutes.
 Assume that the population std is 18.9403.
Can we infer that the mean amount of time
spent by all executives reading and sending
e-mail is different from 60 minutes?
 Assume 5% significance level.
 Calculate the p-value.
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Example 4
 A bank has set up a customer service goal
that the mean waiting time for its customers
will be less than 2 minutes. The bank
randomly samples 30 customers and finds
that the sample mean is 100 seconds.
Assuming that the sample is from a normal
distribution and the standard deviation is 28
seconds, can the bank safely conclude that
the population mean waiting time is less than
2 minutes? Find the p-value.
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Example 5
 A bank has set up a customer service goal
that the mean waiting time for its customers
will be less than 2 minutes. The bank
randomly samples 30 customers and finds
that the sample mean is 112 seconds.
Assuming that the sample is from a normal
distribution and the standard deviation is 28
seconds, can the bank safely conclude that
the population mean waiting time is less than
2 minutes? Find the p-value.
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Answer Key to the Examples Used
 Example 1. H0:  = 60 vs. Ha:  > 60. Rejection
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region: reject H0 if z > 1.645. Given z = 2.48, the
statistical conclusion is to reject H0.
Example 2. H0:  = 60 vs. Ha:  ≠ 60. Rejection
region: reject H0 if z > 1.96 or z < -1.96. Given z =
2.48, the conclusion is to reject H0.
Example 3. 63.6975 ± 1.96 (18.9403/sqrt(162))
Example 1 (cont’d): p-value = 1 – 0.9934 = 0.0066.
Example 2 (cont’d): p-value = 2 × (1 – 0.9934) =
0.0132.
Example 3: p-value ≈ 0.0000.
Example 4: p-value = P( z < -1.56 ) = 0.0594
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