Transcript Starters - GEOCITIES.ws

Unit 6 Starters
Starter 11.1.1
• Draw a standard normal curve on your
calculator
• Set your window to [-3,3]1 by [-.1,.4] .1
• Use y1 = normalpdf(x)
• Draw a neat, reasonably large sketch of
the curve in your notes. Show scales.
• We will use this later
Starter 11.1.2
The heights of 25 students have a mean of
186 cm with standard deviation 3 cm.
1. What is the standard error of the distribution
of sample means?
2. Assuming the population mean height (µ) of
the population is 185, what is the t test
statistic?
3. What is the probability of getting a sample
mean of 186 or higher?
Starter answers
• SE = 3/√25 = 3/5 = .6
186  185
t
 1.67
.6
• P-value from the table
– Row 24
– So:
1.318 < t < 1.711
.10 > P > .05
• P-value from the calculator
– tcdf(1.67,999,24)=0.054
Starter 11.1.3
• Scores on the AP exam are supposed to have a mean of 3.
Mr. McPeak thinks his students score higher than that, so he
takes a sample of ten of last year’s scores
• Here are the scores:
2
3
2
5
5
4
3
1
4
4
• Using the methods we learned yesterday:
– Find a 95% confidence interval for the mean from the formula
– Perform a hypothesis test that might support Mr. McPeak’s claim
• Clearly state your conclusion in a sentence
Starter answers
•
•
•
•
•
From 1-var Stats: Ë = 3.3
S = 1.337
SE = 1.337/sqrt(10) = .4228
t* = 2.262
C.I. = 3.3 ± 0.956 = (2.343, 4.257)
Ho: μ = 3
Ha: μ > 3 Choose α = .05
– Note: Choice of α is arbitrary but reasonable.
• t = (3.3 – 3) / .4228 = .709
• P = tcdf(.709,999,9) = .248
• Conclusion: There is not sufficient
evidence (t = .709, P = .248) to support
the claim that the mean is greater than 3
Starter 11.1.4
• Seven students took an SAT prep course after
doing poorly on the test. Here are their before
(row 1) and after (row 2) scores:
1020 1100 1110
1000 990
1050 1060
1030 1120 1100
1050 1040 1110
1160
• Use a matched-pairs t-test on your calculator
to determine if they improved their scores in a
statistically significant way
Starter solution
• Put “before” in L1 and “after” in L2
• Let L3 = L2 – L1 (or reverse)
• State hypotheses:
– Ho: μ = 0
– Ha: μ > 0
• Stat : Tests : T-Test
– Choose Data, μo = 0, L3, μ > μo
• There is good evidence (t6 = 2.90, p = .01)
to support the claim that the SAT prep
course led to improved scores.
Starter 11.1.5
• Write the important elements that must be present
in any experimental design? What is the desireable
(but not mandatory) element of experimental
design?
• Comparison
– There should be a treatment group and a control group
• Randomization
– Subjects must be randomly assigned to a group
• Replication
– There must be a large enough sample in each group that
the results are statistically significant
• Blindness (optional but highly desirable)
– Subjects should not know which group they are in
– Experiments should not know either (if possible)
Starter 11.2.1
• Write the assumptions that underlie the
use of the t test.
• Which assumption is so important you
can’t work without it?
• What is the best way to tell if the
distribution assumption is met?
Starter Solution
• The two main assumptions we need are:
– The sample came from a valid SRS of the population.
– The population is approximately normally distributed.
• We can’t do anything without a valid SRS.
• Plot the data to see if they are approximately
normal.
– Note that if sample size is large we can live with
skewness or outliers.
Starter 11.2.2
• In the early eighties, a large group of 13 year
olds were given the SAT. Verbal results were
nearly identical, but there was a real
difference in the math results. Here are the
facts:
– 19,883 boys had a mean score of 416 with
standard deviation of 87
– 19,937 girls had a mean score of 386 with a
standard deviation of 74
• Use the formula to state a 99% confidence
interval for the mean difference of scores
between ALL boys and girls.
Starter solution
• From the table, t* = 2.576
87 2
742
416  386  2.576

 30  2.1  (27.9,32.1)
19883 19937
• So we are 99% confident that the true
difference of the boys’ mean score minus
the girls’ mean score is between 27.9 and
32.1
Starter 12.1.1
• Do dogs who are house pets have higher
cholesterol than dogs who live in a
research clinic? A clinic measured the
cholesterol level in all 23 of its dogs and
found a mean level of 174 with s.d. of 44.
They also measured 26 house pets
brought in to be neutered one week and
found a mean of 193 with s.d. of 68.
• Is this strong evidence that house pets
have higher cholesterol than clinic dogs?
• What is wrong with this study?
Solution
• Use the calculator’s 2-sample TTest
• t = 1.174
p = 0.123
• There is not enough evidence to support the
claim that house pets have higher cholesterol
than clinic dogs.
• The problem with this study is that there is not
proper randomization. The clinic used all its
dogs and used pets that happened to be in the
clinic for treatment. This casts serious doubt on
the validity of the conclusion.
Starter 12.1.2
Some people think that chemists are more likely than others to
have female children. Perhaps they are exposed to chemicals
that cause this. Between 1980 and 1990 in Washington state,
555 children were born to chemists. Of these births, 273 were
girls. During this period, 48.8% of all births in Washington were
girls. Is there evidence that the proportion of girls born to
chemists is higher than normal?
Write hypotheses, calculate the sample proportion, perform a
test, and write your conclusion.
Starter solution
• Ho: p = .488
Ha: p > .488
• p-hat = 273 / 555 = .492, so find z and p
.492  .488
z
 .188
(.488)(.512)
555
• p = normalcdf(.188,999) = .43
• There is not good evidence (p = .43) that
the proportion of chemists’ girls is higher
than normal.
Starter 12.2.1
One-sample procedures for proportions can also be used in
matched pairs experiments. Here is an example:
Each of 50 randomly selected subjects tastes two
unmarked cups of coffee and says which he/she prefers.
One cup in each pair contains instant coffee; the other is
fresh-brewed. 31 of the subjects prefer fresh-brewed.
1.
2.
3.
Test the claim that a majority of people prefer the taste of
fresh-brewed coffee. State hypotheses, check assumptions,
find the test statistic and p-value. Is your result significant at
the 5% level?
Find a 90% confidence interval for the true proportion that
prefer fresh-brewed.
When you do an experiment like this, in what order should
you present the two cups of coffee to the subjects?
Starter Solution
• Ho: p = .5
• Assumptions
Ha: p > .5
– SRS? OK
– Large population? All coffee drinkers > 10x50 OK
– At least 10 in each group? 50 x .5 > 10 OK
• z = 1.70 p = .045
• There is sufficient evidence (p<.05) to support
the claim that coffee drinkers prefer freshbrewed.
• C.I. = .62 ± 1.645 x .0686 = (.507, .733)
• Randomization is needed in any experiment;
flip a coin (or use another method) to choose
which coffee each subject gets first.
• See question on next slide:
The Cola Challenge!
• Do Northgate students prefer Pepsi over
Coke? In a randomized matched-pairs
experiment we did last fall, here were the
results:
– Preferred Pepsi: 58
– Preferred Coke: 31
• Perform a hypothesis test of the claim that
Pepsi is preferred.
Starter 12.2.2
The drug AZT is used to treat symptoms of AIDS. It was
studied in an experiment involving volunteers already
diagnosed as having HIV, the virus that causes AIDS. 435
subjects took AZT and another 435 took a placebo. At the
end of the study, 17 of the AZT subjects had developed
AIDS; 38 of the placebo subjects had developed AIDS.
We want to test the claim that taking AZT lowers the
proportion of people who go from HIV to AIDS.
1.
2.
3.
4.
Assign numbers to the groups
Verify that assumptions are met and state hypotheses
Carry out the test on calculator and write your conclusion
This experiment was double-blind. What does that mean?
Starter Solution
• Let Group 1 be the AZT & Group 2 the placebo
• Assumptions
– SRS
– Each population at least 10 times sample size
– Each count of yes or no at least 5
• Ho: p1 = p2
Ha: p1< p2
• Stat:Tests:2-PropZTest yields Z = -2.93 and
p
= .0017
• Conclude there is strong evidence to support the
claim that AZT reduces proportion who get AIDS
• Double Blind: Neither the subject nor the person
administering the drug knows if the subject gets
the AZT or the placebo
Starter 13.1.1
•
Sickle-cell trait is a hereditary condition that is
common among blacks and can cause medical
problems. Some biologists suggest that the
trait protects against malaria. A study in Africa
tested 543 children for the sickle-cell trait and
also for malaria. In all, 136 of the children had
the sickle-cell trait, and 36 of these had heavy
malaria infections. The other 407 children
lacked the trait, and 152 of them had heavy
malaria infections.
1. What are the two populations of interest here?
2. Give a 95% Confidence Interval for the difference in
proportions of malaria in the two populations.
3. Is there good evidence that the proportion of heavy
malaria infection is lower among children with the
sickle-cell trait?
Answer
• The two populations are children with the trait and
children without the trait.
• Use the TI 2-PropZInt screen: (-.197, -.021)
– I am 95% confident that the malaria proportion in children
with sickle-cell is between 2% and 20% less than in
children without the trait.
• Because 0 was not in the confidence interval, there
is good evidence to support the claim that the
sickle-cell trait protects against malaria.
– Note: If you run the 2-PropZTest, z = -2.3, p = .01
Starter 13.1.2
Elite distance runners are thinner than the rest of
us. Skinfold thickness, which indirectly
measures body fat, can show this. A random
sample of 20 runners had a mean skinfold of 7.1
mm with a standard deviation of 1.0 mm. A
random sample of 95 non-runners had a mean
of 20.6 w/ sd of 9.0.
Form a 95% confidence interval for the mean
difference in body fat between runners and nonrunners.
Starter Solution
•
•
•
•
•
•
Choose 2-SampTInt
Enter x1=7.1
Sx1=1
n1=20
Enter x2=20.6 Sx2=9
n2=95
Enter C = .95
Find (-15.38, -11.62)
Conclusion: We are 95% confident that
the true mean skinfold of runners is
between 15.4 mm and 11.6 mm less than
non-runners.
Starter 13.1.3
According to the NCAA, 45 out of 74 athletes
admitted to a certain university in 1994
graduated within 6 years. Assuming this is a
valid sample of all athletes, does the proportion
of athletes who graduate differ from the alluniversity proportion, which is 68%?
State hypotheses, perform a test, write a
conclusion.
Starter Solution
• This is a one-sample proportion test
– Now what are the hypotheses?
• Ho: p = .68
Ha: p ≠ .68
• Use Stat:Tests:1-PropZTest
– po = .68
– z = -1.32
x = 45
p = .185
n = 74
• Conclusion: There is not sufficient evidence
(p = .185) to support a claim that the
graduation rate of athletes differs from nonathletes.
• Is there a different approach that could be
taken to get the same result?
Starter 13.2.1
A study of iron deficiency in infants compared two groups.
One had been breast-fed, the other had been fed formula
from a bottle. The hemoglobin levels were measured at
age 12 months. Here are the results
Group
n
mean
Std dev
Breast-fed
23
13.3
1.7
Bottle-fed
19
12.4
1.8
Assuming this was a properly randomized experiment, is
there significant evidence that the mean hemoglobin level
is different between the groups?
Why is the assumption that we had a properly designed
experiment questionable?
Starter Solution
• This is a two-sample means problem
– Assumptions needed:
• SRS from each population
• Independent populations
• Sample means approximately normally distributed
– Use 2-SampTTest or 2-SampTInt
• Ho: μ1 = μ2
Ha: μ1 ≠ μ2
α = .05
• t = 1.65
p = .107
• There is not sufficient evidence (p = .107) to
support a claim that there is a difference in the
hemoglobin level between the two groups.
Starter 13.2.2
I rolled a die 60 times and got the
following distribution of results:
Outcome
1
2
3
4
5
6
Quantity
6 10 7
11
8
18
Is the die fair? Perform a test and state
your conclusion.
Starter Solution
•
•
•
•
•
•
Observed outcomes in L1 {6, 10, 7, 11, 8, 18}
Expected outcomes in L2 {all 10’s}
(O – E)²/E in L3
Sum L3 to get X² of 9.4
X²cdf(9.4, 999, 5) = .094
Conclusion: There is not sufficient evidence
(p=.09) to support a claim that the die is
unfair.
Dice Day Starter
•
•
Do unregulated providers of child care in their homes
follow different health and safety practices in different
cities? A study looked at people who regularly provided
care for someone else’s children in poor areas of three
cities The numbers who required medical releases from
parents to allow medical care in an emergency were 42
of 73 providers in Newark, 29 of 101 in Camden and 48
of 107 in Chicago.
Is there a significant difference among the proportions
of providers who require medical releases in the three
cities?
1.
2.
3.
4.
Identify the two variables and write a two-way table of counts.
Write the null and alternative hypotheses.
Perform the test and draw a conclusion.
Verify that the necessary conditions are met.
Solution
• Ho: There is no
association between city
and requirement.
• Ha: There is an
association between city
and requirement.
• Run X2 test on calc.
• There is good evidence to
support a claim that
requirements differ by city
• Check expected counts in
matrix [B]
Newark
Camden
Chicago
Require
Not Req.
42
29
48
31
72
59
Starter 14.1.1
The Goodwill second-hand stores did a survey of their customers
in Walnut Creek and Oakland. Among other things, they noted
the sex of each respondent. Here is the breakdown:
Men
Women
W.C.
38
203
Oakland
68
150
Is there a significant difference between the proportion of women
customers in the two stores?
1.
Treat this as a two-sample proportion problem
–
Find the z statistic and p value; draw a conclusion
2.
Do a chi-square test
–
Find X² and the p value; draw a conclusion
–
How does X² relate to z?
Starter Solution
• Ho: p1 = p2
• Ha: p1 ≠ p2
• Use the 2-PropZTest on the TI
– Find z = 3.92 And p = 9.0 x 10-5
– Conclude that there is strong evidence that the proportions differ
• Use the chi-square test on the TI
– Find X² = 15.334 And p = 9.0 x 10-5
– Conclude that there is strong evidence that the proportions differ
• Note that X² is the square of the z statistic
• Conclusion: A two-proportion test can be done with either
the z statistic or with a chi-square test
– The result is the same
Starter 14.1.2
The Goodwill Stores of Walnut Creek and
Oakland also did a breakdown of their shoppers
by income. Here are the results:
Income
($1000’s)
W.C.
Oakland
Under 10
70
62
10 – 20
52
63
20 – 25
69
50
25 – 35
22
19
35 +
28
24
Is there good evidence to believe that the
customers of the two stores have different
income distributions?
Starter Solution
• Ho: There is no difference in the income
distributions
• Ha: The income distributions differ
• Put the two-way table into matrix [A]
• Run the chi-square Test
• X2 = 3.955
p = .412
• Conclusion: There is not sufficient
evidence (p = .412) to support a claim that
the income distributions are different
Starter 14.3.1
Men and women were observed playing a
game of chance. 3 of 12 men won the
game; 8 of 12 women won.
Is there a statistically significant difference
between the men’s and women’s results?
State hypotheses, perform a test and write
a conclusion
Starter solution
• This asks for a comparison of proportions from
two populations
• Use 2-PropZTest screen
– Choose p1 ≠ p2 alternative
– Get z = 2.05 and p = 0.04
• Or use X2 Test screen
– Get X2 = 4.196 and p = .04
• Conclusion:
– There is good evidence (p = .04) that the winning
proportions are different for men and women
– Checking matrix [B] shows all expected counts are at
least 5.