



Study with the several resources on Docsity
Earn points by helping other students or get them with a premium plan
Prepare for your exams
Study with the several resources on Docsity
Earn points to download
Earn points by helping other students or get them with a premium plan
Community
Ask the community for help and clear up your study doubts
Discover the best universities in your country according to Docsity users
Free resources
Download our free guides on studying techniques, anxiety management strategies, and thesis advice from Docsity tutors
This is the Exam Key of Applications of Statistics which includes Variable, Researcher, Type of Variable, Observational Unit, Sample Size, Dendritic Branches, Emanating, Humans and Animals, Disease etc. Key important points are: Definition, Significance Level, Resulting, Hypothesis, Error, Calculate, Confidence Interval, Narrower, Wider, Same Sample
Typology: Exams
1 / 6
This page cannot be seen from the preview
Don't miss anything!
Exam 2 Name ______ANSWER KEY________
Fall 2006
PแผYเตjแฝเต (^) nCj p j^ แบ1โpแป nโj
Yเดฅโฮผ ฯ เตโn
Yเดฅเตt ฮฑ เต 2
s โn
Yเดฅ^ โฮผ s เตโn
แบYเดฅ 1 โYเดฅ 2 แปเต t ฮฑ เต 2
s 12 n 1
s 22 n 2
t เต
เถจ s^1
2 n 1 เต
s 22 n 2
Part I: Answer eight of the following nine questions. If you complete more than eight, I will grade only the first eight. Five points each.
The probability under H 0 of observing a test statistic as extreme or more extreme (in the direction of HA) as that actually observed.
significance level, resulting in a P-value of 0.003. Obviously, in this case, we reject H 0.
If an error was made, it would be a Type I / Type II / neither error.
(Circle the correct answer) Suppose you have calculated a 95% confidence interval for the mean, ฮผ. Now, you want to calculate a 99% confidence interval using the same sample. The 99% confidence interval will be narrower / wider than the 95% confidence interval.
Consider taking a random sample of size 4 from a population of persons who smoke and recording how many of them, if any, have lung cancer. Let pเท represent the proportion of persons in the sample with lung cancer. List the possible values of pเท.
0, 0.25, 0.5, 0.75, 1
(Circle the correct answer) Trichotillomania is a psychiatric illness that causes its victims to have an irresistible compulsion to pull their own hair. Two drugs were compared as treatments for trichotillomania in a study involving 13 women. Each woman took clomipramine during one time period and desipramine during another time period in a double-blind experiment (the women nor their doctors knew which drug they were taking during which time period). We would use the independent / dependent (paired) samples method in order to conduct a test of hypothesis.
The Type II Error rate, ฮฒ = P{failing to reject H 0 |H 0 is false}, for a hypothesis test was calculated to be ฮฒ = 0.07. What is the power = P{rejecting H 0| H 0 is false} for this test?
Power = 1 โ ฮฒ = 1 โ 0.07 = 0.
Part II: Answer every part of the next two problems. Read each question carefully, and show your work for full credit.
1a) (25 pts) A scientist conducted a study to test whether a parakeet chirps more often if there is music playing. The scientist took a random sample of 28 parakeets. Each of the 28 parakeets were observed for two different 30 minute periods โ one with music playing and one without. Using the 0.01 significance level, test whether the mean number of chirps (per 30 minutes) when music is playing is higher than when the room is silent.
Iโve numbered the steps for you on the next page. Please put the appropriate step next to the appropriate number.
(1) ฮฑ = 0.
(2) H0: ฮผm โ ฮผw = 0 or ฮผm = ฮผw or ฮผd = 0 H A: ฮผm โ ฮผw > 0 or ฮผm > ฮผw or ฮผd > 0
TI-84 STAT -> TESTS -> Ttest (using the difference column)
(3) t s = 19.
(4) P = 9.87 x 10 -
(5) P < ฮฑ, so reject H 0
(6) There is significant evidence (at the ฮฑ = 0. significance level) that the true mean number of parakeet chirps (per 30 minutes) is larger when music is playing than when the room is silent.
Parakeet Music No Music Difference 1 12 3 9 2 14 1 13 3 11 2 9 4 13 1 12 5 20 5 15 6 14 3 11 7 10 0 10 8 12 2 10 9 8 6 2 10 13 3 10 11 14 2 12 12 15 4 11 13 12 3 9 14 13 2 11 15 8 0 8 16 18 5 13 17 15 3 12 18 12 2 10 19 17 2 15 20 15 4 11 21 11 3 8 22 22 4 18 23 14 2 12 24 18 4 14 25 15 5 10 26 8 1 7 27 13 2 11 28 16 3 13 Mean 13.7 2.8 10. SD 3.4 1.5 2.
1b) (5 pts) Comment on whether the assumption that can be checked using the QQplot below seems to be met for this analysis.
Notice, our data set is discrete, so it canโt possibly come from a normal distribution. But, we still check the โshapeโ of it with a QQplot, making sure itโs not too different from symmetric and bell-shaped, so that the CLT will give us that the distribution of the sample mean is approximately normal. The points on this QQplot are making a linear pattern. The points that are grouped together horizontally are just repeated observations (this happens frequently with discrete data). There is the one point above the line in the upper tail, but Iโm not too worried about this, since there is no systematic departure from the line. This QQplot shows that the distribution is shaped like a normal, so the moderate sample size of 28 will allow us to use the Central Limit Theorem, meeting our normality assumption.