Statistics 201 – Spring 2021

Exam 3 – Practice Exam (from Spring 2016)

KEY

Disclaimer:

This practice exam is provided solely for the purpose of familiarizing you with the format and style of the Stat 201 exams. There is no explicit or implicit guarantee that the upcoming exam will ask similar questions. If you use the practice exam as your only tool to help you prepare for the upcoming exam, you most likely will not do well on the exam. You should still do the things you would have done if you did not have access to this practice exam, such as re‐read the text, go over your class notes, re‐work the online homework problems, and look at the list of exam topics provided and make sure that you understand all the concepts listed within it.

3. A researcher believes that British men are shorter, on average, than men from theUSA. A random sample of35 men from the USA (ages 20-50) was observed forheight. The mean height was 68.1 inches and the sample standard deviation was 5 .34inches. Meanwhile, a random sample of 35 men from Britain (ages 20-50) wasobserved for height. The mean height was 64.6 and the sample standard deviation was5.81 inches.

i) (2 points) Which type of problem is this? Circle the best answer:

a) One sample proportion testb One sample test of a mean

Two independent samples t-test Chi-Squared test of independence

ii) (2 points) Write down the null and alternative hypotheses suggested by the firstsentence of this problem (HINT: to avoid confusion later, look at the top line of theJMP output below now).

HO: )..,/ V5A- - j) Br-ffq',;1 -:::::: 0

H 4 ! t(J v1;,4 - lJa,;.fc,;11 7 O

iii) (3 points) Below is some output from JMP for a proper analysis of these data.Calculate the missing value (to the right of the word "t Ratio").

eo..s 'I �'J: Ila rd \,1,t?.)i:

t � 3, '-f 1?_! = t�� � 3,� /, "3:5'3f

:::. 2, 62'(

USA-Britain

Assuming unequal variances

Difference 3.49714 t Ratio

Std Err Dif 1.33411 OF

-

67.51277

Upper CL Dif 6.15967 Prob > ltl 0.0108* -5 -3 -1.0054* 1

12 3 4 5Lower CL Dif 0.83461 Prob> t

Confidence 0.95 Prob< t 0.9946

iv) (1 point) There are 3 p-values displayed in the output. Circle the p-value thatcorresponds to your alternative hypothesis in part [ii] above.

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11. Suppose one takes many, many repeated random samples of size n=50 from a verylarge population (of quantitative observations) that is slightly skewed right, has a meanof 127.5 and a standard deviation of 28.5.

i) (2 points) What would be the average of the sampling distribution of the sample mean?

ii) (2 points) What would be the standard deviation of the sampling distribution of thesample mean?

iii) (2 points) Assume all the of necessary conditions are met for the Central LimitTheorem to apply to the sampling distribution of the sample mean (which, they AREmet in this case). Given that the Central Limit Theorem applies, what would be theshape of the sampling distribution of the sample mean?

iv) (6 points) In one particular sample of n=50, a sample mean of 150.7 was obtained.Would you consider this to be a “rare” or “unusual” value of the sample mean? Useyour answers above to help you answer this question. Provide whatever arithmetic youneed to help you answer this question, and be sure to show your work. Circle one:“rare event”Briefly justify the choice you circled above. Limit your answer to two or lesssentences.

(NOTE 1: If you have NO answers to parts [i] and/or [ii], make up answers for thoseparts here, and use that to answer this question. NOTE 2: If you have a calculatorthat can calculate the probability of this event, do NOT use that capability!)

The z-score is well over 3 which indicates this is a very rare event.

Formula Sheet

One Sample Proportions

Confidence interval for p: fJ ± z* t 'fJ-Po Test statistic for a hypothesis test: z = -===

✓Poqo/n

One Sample Means

Confidence interval for µ: - + t* s Y - n-1 -Jn

Test statistic for a hypothesis test:

Difference in means

2 2

Independent samples - CI for µ1 - µ2: (.Y1 - .Y2) ± tvF :: + ::

Independent samples - Test statistic for a hypothesis test: tvF = CYi -y2)-o

DF- Is available from JMP output

Test of Independence

2 S f (obs-Exp)2

XvF = umo Exp where E

Row Total xColumn Total xp = -------­

n

DF = (#Rows - 1) x (#Columns - 1)

s2 sz

... l+i n1 nz

for each cell.

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Sampling Distribution of

µ(y) = µ

y

nSD σ(y) =

SD(y)y

Z y− µ

=