Tests of Significance

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Chapter 11.1 Inference for the Mean of a
Population.
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Example 1 One concern employers have about the
use of technology is the amount of time that
employees spend each day making personal use of
company technology, such as phone, e-mail,
internet, and games. The Associated Press
reports that, on average, workers spend 72
minutes a day on such personal technology uses.
A CEO of a large company wants to know if the
employees of her company are comparable to this
survey. In a random sample of 10 employees, with
the guarantee of anonymity, each reported their
daily personal computer use. The times are
recorded at right.
Employee Time
1 66
2 70
3 75
4 88
5 69
6 71
7 71
8 63
9 89
10 86
When the standard deviation of a statistic is
estimated from the data, the result is called the
standard error of the statistic, and is given by
s/vn.
When we use this estimator, the statistic that
results does not have a normal distribution,
instead it has a new distribution, called the
t-distribution.
Does the data provide evidence that the mean for
this company is greater than 72 minutes?
What is different about this problem?
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Time for some Nspiration!
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One-Sample z-statistic

• s known

m
z
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One-sample t-statistic

• s unknown

m
t
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The variability of the t-statistic is controlled
by the Sample Size. The number of degrees of
freeom is equal to n-1 .
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• ASSUMING NORMALITY?
• SRS is extremely important.
• Check for skewness.
• Check for outliers.
• If necessary, make a cautionary statement.
• In Real-Life, statisticians and researchers try
  very hard to avoid small samples.

Use a Box and Whisker to check.
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Example 2 The Degree of Reading Power (DRP) is a
test of the reading ability of children. Here
are DRP scores for a random sample of 44
third-grade students in a suburban
district 40 26 39 14 42 18 25 43 46 27 19 47 19 2
6 35 34 15 44 40 38 31 46 52 25 35 35 33 29 34 41
49 28 52 47 35 48 22 33 41 51 27 14 54 45 At the
a .1, is there sufficient evidence to suggest
that this districts third graders reading
ability is different than the national mean of 34?
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SRS?

• I have an SRS of third-graders

Normal? How do you know?

• Since the sample size is large, the sampling
  distribution is approximately normally
  distributed
• OR
• Since the histogram is unimodal with no outliers,
  the sampling distribution is approximately
  normally distributed

Name the Test!! One Sample t-test for mean
Do you know s?
What are your hypothesis statements? Is there a
key word?

• s is unknown

Plug values into formula.
p-value tcdf(.6467,1E99,43).2606(2).5212
Use tcdf to calculate p-value.
a .1
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Compare your p-value to a make decision
Conclusion
Since p-value gt a, I fail to reject the null
hypothesis.
There is not sufficient evidence to suggest that
the true mean reading ability of the districts
third-graders is different than the national mean
of 34.
Write conclusion in context in terms of Ha.
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Back to Example 1. The times are recorded
below. Employee 1 2 3 4 5 6
7 8 9 10 Time 66 70 75 88 69 71
71 63 89 86 Does this data provide evidence
that the mean for this company is greater than 72
minutes?
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SRS?

• I have an SRS of employees

• Since the histogram has no outliers and is
  roughly symmetric, the sampling distribution is
  approximately normally distributed

Normal? How do you know?
Do you know s?
What are your hypothesis statements? Is there a
key word?

• s is unknown, therefore we are using a 1 sample
  t-test

H0 m 72 where m is the true of min spent on
PT Ha m 72 time spent by this companys
employees
Use tcdf to calculate p-value.
Plug values into formula.
p-value tcdf(.937,1E99,9).1866(2).3732
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Compare your p-value to a make decision
Conclusion
Since p-value gt 15, I fail to reject the null
hypothesis that this companys employees spend 72
minutes on average on Personal Technology uses.
There is not sufficient evidence to suggest that
the true amount of time spent on personal
technology use by employees of this company is
more than the national mean of 72 min.
Write conclusion in context in terms of Ha.
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Now for the fun calculator stuff!
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Example 3 The Wall Street Journal (January 27,
1994) reported that based on sales in a chain of
Midwestern grocery stores, Presidents Choice
Chocolate Chip Cookies were selling at a mean
rate of 1323 per week. Suppose a random sample
of 30 weeks in 1995 in the same stores showed
that the cookies were selling at the average rate
of 1208 with standard deviation of 275. Does
this indicate that the sales of the cookies is
different from the earlier figure?
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• Assume
• Have an SRS of weeks
• Distribution of sales is approximately normal due
  to large sample size
• s unknown
• H0 m 1323 where m is the true mean cookie
  sales
• Ha m ? 1323 per week
• Since p-value lt a of 0.05, I reject the null
  hypothesis. There is sufficient to suggest that
  the sales of cookies are different from the
  earlier figure.

Name the Test!! One Sample t-test for mean
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• Example 3 Presidents Choice Chocolate Chip
  Cookies were selling at a mean rate of 1323 per
  week. Suppose a random sample of 30 weeks in
  1995 in the same stores showed that the cookies
  were selling at the average rate of 1208 with
  standard deviation of 275. Compute a 95
  confidence interval for the mean weekly sales
  rate.
• CI (1105.30, 1310.70)
• Based on this interval, is the mean weekly sales
  rate statistically different from the reported
  1323?

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What do you notice about the decision from the
confidence interval the hypothesis test?
Remember your, p-value .01475 At a .02, we
would reject H0.

• What decision would you make on Example 3 if a
  .01?
• What confidence level would be correct to use?
• Does that confidence interval provide the same
  decision?
• If Ha m lt 1323, what decision would the
  hypothesis test give at a .02?
• Now, what confidence level is appropriate for
  this alternative hypothesis?

A 96 CI (1100, 1316). Since 1323 is not in
the interval, we would reject H0.
You would fail to reject H0 since the p-value gt a.
You should use a 99 confidence level for a
two-sided hypothesis test at a .01.
The 98 CI (1084.40, 1331.60) - Since 1323
is in the interval, we would fail to reject
H0. Why are we getting different answers?
Tail probabilities between the significant level
(a) and the confidence level MUST match!)
In a CI, the tails have equal area so there
should also be 2 in the upper tail
CI (1068.6 , 1346.40) - Since 1323 is in
this interval we would fail to reject H0.
a .02
.02
.96
That leaves 96 in the middle that should be
your confidence level
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Ex4 The times of first sprinkler activation
(seconds) for a series of fire-prevention
sprinklers were as follows 27
41 22 27 23 35 30 33 24 27 28 22 24 Construct a
95 confidence interval for the mean activation
time for the sprinklers.
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Matched Pairs Test

• A special type of
• t-inference

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Matched Pairs two forms

• Pair individuals by certain characteristics
• Randomly select treatment for individual A
• Individual B is assigned to other treatment
• Assignment of B is dependent on assignment of A

• Individual persons or items receive both
  treatments
• Order of treatments are randomly assigned or
  before after measurements are taken
• The two measures are dependent on the individual

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Is this an example of matched pairs?

• 1)A college wants to see if theres a difference
  in time it took last years class to find a
  job after graduation and the time it took the
  class from five years ago to find work after
  graduation. Researchers take a random sample
  from both classes and measure the number of days
  between graduation and first day of employment

No, there is no pairing of individuals, you have
two independent samples
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Is this an example of matched pairs?

• 2) In a taste test, a researcher asks people in a
  random sample to taste a certain brand of spring
  water and rate it. Another random sample of
  people is asked to taste a different brand
  of water and rate it. The researcher wants to
  compare these samples

No, there is no pairing of individuals, you have
two independent samples If you would have the
same people taste both brands in random order,
then it would be an example of matched pairs.
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Is this an example of matched pairs?

• 3) A pharmaceutical company wants to test its new
  weight-loss drug. Before giving the drug to a
  random sample, company researchers take a weight
  measurement on each person. After a month
  of using the drug, each persons weight is
  measured again.

Yes, you have two measurements that are dependent
on each individual.
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A whale-watching company noticed that many
customers wanted to know whether it was better to
book an excursion in the morning or the
afternoon. To test this question, the company
collected the following data on 15 randomly
selected days over the past month. (Note
days were not consecutive.)
You may subtract either way just be careful
when writing Ha
Day 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15
Morning 8 9 7 9 10 13 10 8 2 5 7 7 6 8 7
After-noon 8 10 9 8 9 11 8 10 4 7 8 9 6 6 9
Since you have two values for each day, they are
dependent on the day making this data matched
pairs
First, you must find the differences for each day.
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Day 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15
Morning 8 9 7 9 10 13 10 8 2 5 7 7 6 8 7
After-noon 8 10 9 8 9 11 8 10 4 7 8 9 6 6 9
Differences 0 -1 -2 1 1 2 2 -2 -2 -2 -1 -2 0 2 -2
I subtracted Morning afternoon You could
subtract the other way!

• Assumptions
• Have an SRS of days for whale-watching
• s unknown
• Since the boxplot doesnt show any outliers, we
  can assume the distribution is approximately
  normal.

You need to state assumptions using the
differences!
Notice the skewness of the boxplot, however, with
no outliers, we can still assume normality!
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Differences 0 -1 -2 1 1 2 2 -2 -2 -2 -1 -2 0 2 -2
Is there sufficient evidence that more whales are
sighted in the afternoon?
Be careful writing your Ha! Think about how you
subtracted M-A If afternoon is more should the
differences be or -? Dont look at numbers!!!!
If you subtract afternoon morning then Ha mDgt0
H0 mD 0 Ha mD lt 0 Where mD is the true mean
difference in whale sightings from morning minus
afternoon
Notice we used mD for differences it equals 0
since the null should be that there is NO
difference.
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Differences 0 -1 -2 1 1 2 2 -2 -2 -2 -1 -2 0 2 -2
finishing the hypothesis test Since p-value
gt a, I fail to reject H0. There is insufficient
evidence to suggest that more whales are sighted
in the afternoon than in the morning.
In your calculator, perform a t-test using the
differences (L3)
Notice that if you subtracted A-M, then your test
statistic t .945, but p-value would be the
same
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Ex The effect of exercise on the amount of
lactic acid in the blood was examined in journal
Research Quarterly for Exercise and Sport. Eight
males were selected at random from those
attending a week-long training camp. Blood
lactate levels were measured before and after
playing 3 games of racquetball, as shown in the
table.
Player Before After
1 13 18
2 20 37
3 17 40
4 13 35
5 13 30
6 16 20
7 15 33
8 16 19
What is the parameter of interest in this
problem? Construct a 95 confidence interval for
the mean change in blood lactate level.
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Based on the data, would you conclude that there
is a significant difference, at the 5 level,
that the mean difference in blood lactate level
was over 10 points?
Player Before After
1 13 18
2 20 37
3 17 40
4 13 35
5 13 30
6 16 20
7 15 33
8 16 19