Before looking at these solutions, please work the practice quiz.
Disclaimer: This quiz is representative of the level of difficulty you should expect, but it doesn’t include every single topic from the week’s work. The real quiz may include some other topics and may skip some that are in this practice quiz. (The real quiz also may not word questions in the same way as the practice quiz. You should focus on the concepts, not a particular form of words.)
See also: How to Take a Math Test
These solutions show about the same level of work I expect from you, though I often add some extra commentary. Please see Show Your Work for the what, why, and how.
See also: Top Ten Mistakes of Hypothesis Tests
1(points: 4) A researcher wanted to see whether the English like soccer more than the Scots. She asked eight randomly selected English and eight Scots to rate their liking for soccer on a numeric scale of 1 (hate) to 10 (love), and she recorded these responses:
From the above data, can the researcher prove that the English have a stronger liking for soccer than the Scots? Use α = 0.05.
Solution: This is numeric data because respondents gave their responses as numbers and it would make sense to average those numbers. The population parameter of concern is μ.
There are two populations, English and Scots. Each individual gives a single number, so the data are unpaired. This is Case 4, Difference of Population Means.
I’ll define pop. 1 as the English, but you could do it the other way as long as you follow through consistently.
Common mistake: You can’t analyze this as Case 3 with paired data. The English and Scots are two independent samples; you’re not getting two numbers from each individual. Homework problem 12–13 was similar.
|(1)|| pop. 1 = English; pop. 2 = Scots.
H0: μ1–μ2 = 0 or μ1 = μ2, the English don’t have a stronger liking than the Scots
H1: μ1–μ2 > 0 or μ1 > μ2, the English have a stronger liking for soccer than the Scots
Remark: You need to show which is population 1 and which is population 2. An acceptable alternative is to use meaningful subscripts, such as E and S, instead of 1 and 2.
|(2)||α = 0.05|
|(RC)||Random samples: given.
These are small samples, so you need to check
that each one is normal (MATH200A part 4)
and that it has no outliers (MATH200A part 2).
Here are the plots, first the normality check for the
English and then for the Scots, followed by a box-whisker for both the
English and the Scots:
The normal probability plots are roughly linear: the r’s of .9734 and .9772 are greater than CRIT = .9054. The boxplots show no outliers.
|(3–4)|| Put the English numbers in L1 and the Scottish numbers in
L2. Perform a 2-SampTTest with Data; L1, L2, 1, 1,
Outputs: t=1.58, p = .0690, df=13.4634, x̅1=6.5375, x̅2=4.8500, s1=1.9146, s2=2.3440
Remark: The means and standard deviations are new information, and you need to write them down. This also helps you figure out was wrong if you get a wrong p-value.
|(5)||p > α: fail to reject H0.|
|(6)|| At the 0.05 level of significance,
we can’t say whether English or Scots have a stronger
liking for soccer.
Remark: Remember, neutral language on a “fail to reject H0”.
2(points: 4) Another researcher took a different approach. She polled random samples with the question “Do you watch football at least once a week?” (In the UK they call soccer “football”). She got these results:
|Sample Size||Number of “Yes”|
At the 0.05 level of significance, are the English and the Scots equally fans of soccer?
Solution: This is attribute data with two responses: a person is either a fan or not. The population parameter of interest is p, the population proportion; and there are two populations, English and Scots. Therefore this is Case 5, Difference in Population Proportions.
Let’s label the English as pop. 1. (If you did it the other way, your hypotheses and your p-value would still be the same as mine, but your test statistic z would have the opposite sign.)
|(1)|| pop. 1 = English pop. 2 = Scots
H0: p1–p2 = 0 or p1 = p2, the English and the Scots have equal proportions of soccer fans
H1: p1–p2 ≠ 0 or p1 ≠ p2, the English and the Scots have different proportions of soccer fans
|(2)||α = 0.05|
|(3–4)|| 2-PropZTest with x1=105, n1=150, x2=160, n2=200, p1≠p2
Output: z = −2.16, p = .0308, p̂1 = 0.70, p̂2 = 0.80, p̂ = 0.7571428751
|(RC)|| (n1+n2)p̂(1−p̂) =
64 ≥ 10, OK.
Are the samples too big? 20n1 = 20×150 = 3000 and 20n2 = 20×200 = 4000. You don’t know the populations of England and Scotland exactly, but they are in the millions and well above 3000 or 4000.
The samples were stated to be random. Therefore the requirements are met, and you can proceed to the conclusion.
|(5)||p < α; reject H0 and accept H1.|
The English and Scots are not equally likely to be soccer fans, at the 0.05 level of significance;
in fact the English are less likely to be soccer fans.
The English and Scots are not equally likely to be soccer fans, (p = .0308); in fact the English are less likely to be soccer fans.
Remark: The problem asked whether the two populations are equally fans; that’s why H1 is ≠, not >. However, if you do reach a significant result in a two-tailed test, you can and should go further and report a population effect in the same direction as the sample effect. Since the English sample had only 70% soccer fans against 80% of the Scots, and the hypothesis test showed a real difference in the population, you conclude that the English population has a lower proportion of soccer fans than the Scots. See p < α in Two-Tailed Test: What Does It Tell You?
3(points: 4) To see if running raises the HDL (“good”) cholesterol level, five female volunteers were randomly selected. They had their HDL level measured before they started running and again after each had run regularly an average of four miles daily for six months. See if you can prove that the average person’s HDL cholesterol level would be raised after all that running. Use α=0.05.
Solution: Each person gave two numeric responses, HDL before running and HDL after running. Therefore this is Case 3, paired numeric data. The population parameter is μd.
Since you expect After to be higher than Before, define d = After − Before. (If you define d as Before−After, everything will work but the calculator will give t and x̅ as negatives.)
|(1)|| d = After−Before
H0: μd = 0, Running doesn’t raise HDL level
H1: μd > 0, Running raises HDL level
Common mistake: It’s not correct to make a hypothesis about μA−μB or μ1−μ2. That would be appropriate for independent samples (unpaired data). But here you have dependent samples (paired data), and μd is the population parameter.
Remark: If this was a research study, they would probably test for a difference in HDL, not just a reduction. Maybe this study was done by a fitness center or a running-shoe company. They would want to find a reduction, and HDL increasing or staying the same would be equally uninteresting to them.
|(2)||α = 0.05|
Random sample: given.
This is a small sample, so use a
box-whisker plot (MATH200A part 2) to check the
outliers, and MATH200A part 4
to check that the d’s come from a
Put Before in L1, After in L2. Set
L3=L2−L1 and use L3 for your tests.
(When there are only a few points, vertical differences are exaggerated in the normal probability plot. The correlation coefficient of 0.9131 is still greater than CRIT = .8804.)
T-Test 0, L3, 1, μ>μo
Outputs: t = 3.06, p = 0.0188 d̅=4.6, s=3.3615..., n=5,
|(5)||p < α; reject H0 and accept H1.|
|(6)||At the 0.05 level of significance, HDL level is higher after running 4 miles daily for six months.|
Solution: You already have the data in your calculator and have already checked the requirements, so all you have to do is compute a TInterval with C-Level = .9. The output is (1.3951, 7.8049).
Interpretation: You are 90% confident that running an average of four miles a day for six months will raise the average woman’s HDL by 1.4 to 7.8 points.
Remark: Notice the correspondence between hypothesis test and confidence interval. The one-tailed HT at α = 0.05 is equivalent to a two-tailed HT at α = 0.10, and the complement of that is a CI at 1−α = 0.90 or a 90% confidence level. Since the HT did find a statistically significant effect, you know that the CI will not include 0; if the HT had failed to find a significant effect then the CI would have included 0. See Hypothesis Test by Confidence Interval.
Solution: Use MATH200A part 5. Since you have no prior estimates, use p̂1 = p̂2 = 0.5. Enter
C-Level: .95, Data Type: 4:2 pop binomial, E: .03, p̂1: .5, p̂2: .5
Read off the answer of 2135 per sample. Answer: she needs at least 2135 English and at least 2135 Scots.
(a) If there is something that you are motivated to prove in a hypothesis test, what you wish to prove must be stated in the [ alternative ] hypothesis.
(b) The [ p-value ] is the actual computed probability of a Type I error (which can occur when you reject H0 and accept H1) for a particular experiment.