Unit 6 Starters

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Unit 6 Starters
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Starter 11.1.1

• Draw a standard normal curve on your calculator
• Set your window to -3,31 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

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Starter 11.1.2

• The heights of 25 students have a mean of 186
  cm with standard deviation 3 cm.
• What is the standard error of the distribution of
  sample means?
• Assuming the population mean height (µ) of the
  population is 185, what is the t test statistic?
• What is the probability of getting a sample mean
  of 186 or higher?

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Starter answers

• SE 3/v25 3/5 .6
• P-value from the table
• Row 24 1.318 lt t lt 1.711
• So .10 gt P gt .05
• P-value from the calculator
• tcdf(1.67,999,24)0.054

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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 years scores
• Here are the scores
• 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.
  McPeaks claim
• Clearly state your conclusion in a sentence

2 2 5 3 4
3 5 4 1 4
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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 gt 3 Choose a .05
• Note Choice of a 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

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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
• Use a matched-pairs t-test on your calculator to
  determine if they improved their scores in a
  statistically significant way

1020 1100 1110 1000 990 1050 1060
1030 1120 1100 1050 1040 1110 1160
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Starter solution

• Put before in L1 and after in L2
• Let L3 L2 L1 (or reverse)
• State hypotheses
• Ho µ 0
• Ha µ gt 0
• Stat Tests T-Test
• Choose Data, µo 0, L3, µ gt µo
• There is good evidence (t6 2.90, p .01) to
  support the claim that the SAT prep course led to
  improved scores.

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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)

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Starter 11.2.1

• Write the assumptions that underlie the use of
  the t test.
• Which assumption is so important you cant work
  without it?
• What is the best way to tell if the distribution
  assumption is met?

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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 cant 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.

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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.

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Starter solution

• From the table, t 2.576
• 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

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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?

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Solution

• Use the calculators 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.

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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.

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Starter solution

• Ho p .488 Ha p gt .488
• p-hat 273 / 555 .492, so find z and p
• p normalcdf(.188,999) .43
• There is not good evidence (p .43) that the
  proportion of chemists girls is higher than
  normal.

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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.
• 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?

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Starter Solution

• Ho p .5 Ha p gt .5
• Assumptions
• SRS? OK
• Large population? All coffee drinkers gt 10x50
  OK
• At least 10 in each group? 50 x .5 gt 10 OK
• z 1.70 p .045
• There is sufficient evidence (plt.05) to support
  the claim that coffee drinkers prefer
  fresh-brewed.
• 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

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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.

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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.
• 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?

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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 p1lt p2
• StatTests2-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

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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.
• What are the two populations of interest here?
• Give a 95 Confidence Interval for the difference
  in proportions of malaria in the two populations.
• Is there good evidence that the proportion of
  heavy malaria infection is lower among children
  with the sickle-cell trait?

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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

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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
  non-runners.

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Starter Solution

• Choose 2-SampTInt
• Enter x17.1 Sx11 n120
• Enter x220.6 Sx29 n295
• 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.

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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
  all-university proportion, which is 68?
• State hypotheses, perform a test, write a
  conclusion.

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Starter Solution

• This is a one-sample proportion test
• Now what are the hypotheses?
• Ho p .68 Ha p ? .68
• Use StatTests1-PropZTest
• po .68 x 45 n 74
• z -1.32 p .185
• Conclusion There is not sufficient evidence (p
  .185) to support a claim that the graduation
  rate of athletes differs from non-athletes.
• Is there a different approach that could be taken
  to get the same result?

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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
• 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?

Group n mean Std dev
Breast-fed 23 13.3 1.7
Bottle-fed 19 12.4 1.8
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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 a .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.

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Starter 13.2.2

• I rolled a die 60 times and got the following
  distribution of results
• Is the die fair? Perform a test and state your
  conclusion.

Outcome 1 2 3 4 5 6
Quantity 6 10 7 11 8 18
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Starter Solution

• Observed outcomes in L1 6, 10, 7, 11, 8, 18
• Expected outcomes in L2 all 10s
• (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.

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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
  elses 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?
• 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.

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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

Require Not Req.
Newark 42 31
Camden 29 72
Chicago 48 59
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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
• Is there a significant difference between the
  proportion of women customers in the two stores?
• Treat this as a two-sample proportion problem
• Find the z statistic and p value draw a
  conclusion
• Do a chi-square test
• Find X² and the p value draw a conclusion
• How does X² relate to z?

Men Women
W.C. 38 203
Oakland 68 150
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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

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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
• Is there good evidence to believe that the
  customers of the two stores have different income
  distributions?

Income (1000s) W.C. Oakland
Under 10 70 62
10 20 52 63
20 25 69 50
25 35 22 19
35 28 24
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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

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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 mens and womens results?
• State hypotheses, perform a test and write a
  conclusion

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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.