Transcript Power point 2

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The Sampling Distribution of a Difference Between Two Means!
Read p. 628 - 632
What are the shape, center, and spread of the sampling distribution of x1  x 2? Are
there any conditions that need to be met?
Shape: If the population distributions are Normal, the sampling distributions of x1  x 2
is Normal. In other cases, the sampling distribution of x1  x 2 will be approximately
Normal if the sample sizes are large enough (n1 > 30 and n2 > 30).
Center: The mean of the sampling distribution is μ1 – μ2.
Spread: The standard deviation of the sampling distribution of x1  x 2 is
as long as each sample is no more than 10% of the population.
 
2
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n1
n2
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Example:"
A potato chip manufacturer buys potatoes from two different suppliers, Riderwood
Farms and Camberley,Inc. The weights of potatoes from Riderwood Farms are
approximately Normally distributed with a mean of 175 grams and a standard
deviation of 25 grams. The weights of potatoes from Camberley are approximately
Normally distributed with a mean of 180 grams and a standard deviation of 30 grams.
When shipments arrive at the factory, inspectors randomly select a sample of 20
potatoes from each shipment and weigh them. They are surprised when the average
weight of the potatoes in the sample from Riderwood Farms is higher than the
average weight of the potatoes in the sample from Camberley x .
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(a) Describe the shape, center, and spread of the sampling distribution of xc  x r.
The shape of the sampling distribution of x c  x r is approximately Normal since both
population distributions are Normal. The mean of the sampling distribution is
180-175=5 grams, and its standard deviation is 252 302
20
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20
 8.73 grams
(b) Interpret the standard deviation of the sampling distribution of xc  x r .
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(c) Find the probability that the mean weight of the Riderwood sample is larger than the mean weight of
the Camberly sample. Should the inspectors have been surprised?
If the mean of the Riderwood sample is larger, then xc  x r must be negative. The graph below shows
the desired probability shaded.
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0 5
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P x c  x r  0  P z 
  P( z  0.57)  0.28
8.73 
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The inspectors shouldn’t be surprised, because the Riderwood sample will have a higher mean more
than one-fourth of the time.
-3.7
-12.5
5
13.7
22.5
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The Two-Sample t Statistic!
Read p.633 - 634
What is the standard error of x1  x 2 ? Is this on the formula sheet?
The formula for standard deviation is but the formula
2
2
s1  s2
for standard error is not.
n1 n2
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What is the formula for the two-sample t statistic? Is this one on the formula sheet?
What does it measure?
x1  x 2  1   2 
t

No it is not.
2
2
s1  s2
It tells us how far x1  x 2 is from its mean in standard
n1 n2
deviations units.
What distribution does the two-sample t statistic have? Why is this a t statistic rather
than a z statistic? How do you calculate the degrees of freedom?
It has approximately a t distribution. Use is the t statistic when we don’t know the
population standard deviation.
There are 2 ways.
1) (technology) There is a some what messy formula.
2) (conservative) Use the t distribution with degrees of freedom equal to the smaller of
n1-1 and n2-1. With this option, the resulting confidence interval has a margin of error
as large as or larger than is needed for desired confidence level. The significance test
using this option gives a P-value greater of equal to the true P-value.
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Confidence Intervals for μ1 - μ2!
Read p.634 - 637
What are the conditions for a two-sample t interval for μ1 – μ2?
Random, Normal, Independent
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What is the formula for the two-sample t interval for μ1 – μ2? Is this formula on the
formula sheet?
2
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This specific formula is not on the formula sheet,
x1  x 2  t * s1  s2
n1 n2
but you are given the general formula for a confidence interval
and the formula for standard deviation of the sampling distribution.
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Is it OK to use your calculator for the Do step? Are there any drawbacks?
It is ok to use your calculators. If you do, do not write the equation using conservative
degrees of freedom. These will not produce the same answer. If you want to write out
the formula, leave t* in the formula instead of putting in the conservative critical value.
Then report the calculator’s interval and degrees of freedom.
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Example:
Do plastic bags from Target or plastic bags from Bashas hold more weight? A group of
AP Statistics students decided to investigate by filling a random sample of 5 bags from
each store with common grocery items until the bags ripped. Then they weighed the
contents of items in each bag to determine its capacity. Here are their results, in grams:
Target
12,572
13,999
11,215
15,447
10,896
Bashas
9,552
10,896
6,983
8,767
9,972
(a) Construct and interpret a 99% confidence interval for the difference in mean capacity
of plastic grocery bags from Target and Bashas.
State: We want to estimate μT – μB at the 90% confidence level, where μT = the mean
capacity of plastic bags from Target (in grams) and μB = the mean capacity of plastic
bags from Bashas (in grams).
Plan: If the conditions are met, we will calculate a two-sample t interval for μT – μB.
Random: The students selected a random sample of bags from each store.
Normal: Since the sample sizes are small, we must graph the data to see if it is
reasonable to assume that the population distributions are approximately Normal.
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Since there is no skewness or outliers, it is safe to use t procedures.
Independent: The samples were selected independently, and it is reasonable to
assume that there are at least 10(5) = 50 plastic grocery bags at each store.
Do: For these data, x T  12,825, sT  1912.5, x B  9234, s B  1474.2. Using a conservative df
of 5-1=4, the critical value for 99% is t* = 4.604. The confidence interval is
1474.2 2 1912.52
12,826  9234  4.604
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 3592  4972  (1380,8564)
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With technology and df = 7.5,
CI = (-101, 7285)
Conclude: We are 99% confident that the interval from -1380 to 8564 grams captures
the true difference in the mean capacity of plastic grocery bags from Target and from
Bashas.
(b) Does your interval provide convincing evidence that there is a difference in the
mean capacity between the two stores?
Since the interval includes 0, it is plausible that there is no difference in the two
means. Thus, we do not have convincing evidence that there is a difference in mean
capacity. However, if we increased the sample size, we would likely find a convincing
difference since it seems pretty clear that Target bags have a bigger capacity.
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Significance Tests for μ1 - μ2"
Read p. 638 - 643
What are the conditions for conducting a two-sample t test for μ1 – μ2?
Random, Normal, and independent
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Example:
In commercials for Bounty paper towels, the manufacturer claims that they are the “quicker pickerupper.” But are they also the stronger picker upper? Two AP Statistics students, Wesley and
Maverick, decided to find out. They selected a random sample of 30 Bounty paper towels and a
random sample of 30 generic paper towels and measured their strength when wet. To do this, they
uniformly soaked each paper towel with 4 ounces of water, held two opposite edges of the paper
towel, and counted how many quarters each paper towel could hold until ripping, alternating
brands. Here are their results:
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Bounty:
106, 111, 106, 120, 103, 112, 115, 125, 116, 120, 126, 125, 116, 117, 114
118, 126, 120, 115, 116, 121, 113, 111, 128, 124, 125, 127, 123, 115, 114
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Generic:
77, 103, 89, 79, 88, 86, 100, 90, 81, 84, 84, 96, 87, 79, 90
86, 88, 81, 91, 94, 90, 89, 85, 83, 89, 84, 90, 100, 94, 87
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(a) Display these distributions using parallel boxplots and briefly compare these distributions.
Based only on the boxplots, discuss whether or not you think the mean for Bounty is significantly
higher than the mean for the generic.
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(b) Use a significance test to determine if there is convincing evidence that wet Bounty paper
towels can hold more weight, on average, than wet generic paper towels.
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a) Display these distributions using parallel boxplots and briefly compare these
distributions. Based only on the boxplots, discuss whether or not you think the mean for
Bounty is significantly higher than the mean for the generic.
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The five-number summary for Bounty paper towels is (103, 114, 116.5, 124, 128), and
the five-number summary for the generic paper towels is (77, 84, 88, 90, 103). Both
distributions are roughly symmetric, but the generic brand has two high outliers. The
center of the Bounty is much higher than the center of the generic distribution. Although
the range of each distribution is roughly the same, the interquartilerange of the Bounty
is larger.
Since the centers are so far apart and there is no overlap in the two distributions, the
Bounty mean is almost certain to be significantly higher than the generic mean. If the
true means were really the same, it would be virtually impossible to get so little overlap.
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(b) Use a significance test to determine if there is convincing evidence that wet Bounty
paper towels can hold more weight, on average, than wet generic paper towels.
State: We want to perform a test of H0: = µB - µG= 0 versus Ha: µB - µG > 0 at a 5%
level of significance, where µB = the mean number of quarters a wet Bounty paper
towel can hold and µG = the mean number of quarters a wet Generic paper towel can
hold.
Plan: If the conditions are met, we will conduct a two-sample t test for µB - µG.
Random: The students used a random sample of paper towels.
Normal: Even though there are two outliers in the generic distribution, both
distributions are reasonably symmetric, and the sample sizes are both at least 30, so it
is safe to use t procedures.
Independent: The samples were selected independently, and it is reasonable to
assume there are at least 10(30) = 300 paper towels of each brand.
Do: For these data, xB  117, S B  6.64, xG  88.1, and SG  6.30
test statistic: t  (117.6  88.1)  0  17.64
6.64 2 6.30 2
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30
30
P-value: Using either the conservative df = 30 – 1 = 29 or from technology, df = 57.8,
the P-value is approximately 0.
Conclude: Since the P-value is less than 0.05, we reject H . There is very convincing
evidence that wet Bounty paper towels can hold more weight, on average, than wet
generic paper towels.
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(c) Interpret the P-value from (b) in the context of this question.
Since the P-value is approximately 0, it is almost impossible to get a difference in
means of at least 29.5 quarters by random chance if the two brands of paper towels
can hold the same mean amount when wet.
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Using Two-Sample t Procedures Wisely"
Read p. 644 - 648
When doing two-sample t procedures, should we pool the data to estimate a common
standard deviation? Is there any benefit? Are there any risks?
Do not pool the data for t procedures. We DO have to pool when doing a significance
test for the difference of 2 proportions but not with means.
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Should you use two-sample t procedures with paired data? Why not? How can you know
which procedure to use?
• Do not use these procedures with paired data because in paired data you arent looking
for the difference between the means of the two groups - you are looking at the mean
of the differences. You determine which process to use by looking at the way the data
is collected... Matched Pairs Experiment = Paired Data = one - sample t test
Completely randomized experiment or stratified sampling = Difference of 2 Means =
two-sample test
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Example:
Suppose you are designing an experiment to determine if students perform better on
tests when there are no distractions, such as a teacher talking on the phone. You have
access to two classrooms and 30 volunteers who are willing to participate in your
experiment.
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(a) Design an experiment so that a two-sample t test would be the appropriate
inference method.
On 15 index cards write “A” and on 15 index cards write “B”. Shuffle the cards and
hand them out at random to 30 volunteers. All 30 subjects will take the same reading
comprehension test. Subjects who receive A cards will go to a classroom with no
distractions, and subjects who receive B cards will go to a classroom that will have the
proctor talking on the phone during the test. At then end of the experiment, compare
the mean score for subjects in room A with the mean score for subjects in room B.
(b) Design an experiment so that a paired t test would be the appropriate inference
method.
Using the same procedure in part (a), divide the subjects into two rooms and give them
the same reading comprehension test. One room will be distraction free, and the other
room will have a proctor talking on the phone. Then, after a short break, give all 30
subjects a similar reading comprehension test but have the distraction in the opposite
room. At the end of the experiment, calculate the difference in the two reading
comprehension scores for each subject and c9ompare the mean difference with 0.
(c) Which experiment design is better? Explain.
The experiment design in part (b) is better since it eliminates an important source of
variability – the reading comprehension skills of the individual subjects.
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