STP 226

Instructor: Ela Jackiewicz

Example EXAM #2 Material from chapters 7-9

Honor Statement:I have neither given nor received information regarding this exam, and I will not do so until all exams have been graded and returned.

Signed ______________________

Date_________________________

______________________________ ___________________________PRINTED NAME CLASS TIME

DIRECTIONS:

This is a closed book examination. You may use one page 8X11 (one side only) with hand written notes and a graphing calculator. Formulas , z-tables and t-tables are included at the end of the test. For the first 5 questions you need to provide complete and well-organized answers, Include sketches as requested to explain your answers, Round up all answers to required decimal places. For multiple choice questions #6-#14 place letter answers in the tables below. You can earn 105 points (5 points are extra credit points ).

RELAX and Good Luck!

Place answers to multiple choice questions below, use capital letters A-E as appropriate:

Question number #6 #7 #8 #9 #10 #11 #12 #13

Answer:

Question number

(extra credit)

#14

Answer:

Part one: Show all work for questions #1-# 5

Use following information for Questions #1 and #2 In the library on a university campus, there is a sign in the elevator that indicates a limit of 32 persons. Furthermore, there is a weight limit of 5000 lb. Assume that average

weight of students, faculty, and staff on campus is μ =150 lb, that standard deviation

of weights is σ =27 lb. Use Central Limit Theorem to answer Questions #1 and #2

Question #1 (6 points)

Suppose samples of 32 persons from the campus are to be taken and x̄ is to be their

average weight. What is the sampling distribution of x̄ ? Give the mean and the standard deviation of that distribution.

Distribution is :______________________________________

μ x̄ =______________________ σ X̄ =______________

Question #2 (6 points)What is the probability that for a random sample of 32 persons on the elevator their

average weight x̄ will be greater than 156.25 lb, so that the total weight will exceed the weight limit of 5000 lb? Show work, include appropriate sketch explaining your answer.

P(X̄>156.25)= __________________

Use following information in Questions #3-#5

A hot dog manufacturer claims that one of his brands of hot dogs called So-Lean has an

average fat content less than 18g per hot dog. Let μ be the true mean fat content perhot dog in all the hotdogs of So-Lean brand. An independent testing organization is asked to analyze a random sample of 36 So-Lean hot dogs to assess if manufacturer's

claim is correct. Let x̄ be an average fat content for this sample.

Question#3 (8 points)

Suppose that the sample resulted in a mean x̄ =18.6 g and standard deviation s=1g.

Compute 95% confidence interval for μ = mean fat content per hot dog in all the hotdogs of that brand. Use a proper procedure. Clearly show all work by hand.

Question#4 (8 points)

Suppose 99% Confidence Interval for μ = mean fat content per hot dog in all the hotdogs of that brand is (18.1 g, 19.1 g). Based on that interval, do you think that manufacturer's claim is correct? Clearly write a short explanation of your decision.

Correct Not Correct (circle one)

Explanation:__________________________________________________

Question#5 (16 points)

Suppose that the random sample of 36 So-Lean hot dogs resulted in a mean x̄ =17.6 gand sample standard deviation was 1.7g. Test the following hypothesis:

H 0:μ=18 against Ha :μ<18 at 5% significance level. Use following parts clearly showing all work:

a) Compute the appropriate test statistics.

b)Give the rejection region for your test, include appropriate sketch, marking critical value(s), and clearly label rejection and non-rejection areas.

c)Decide if null hypothesis is rejected or not, explain your decision

Reject H 0 Do not reject H 0 (circle one)

Explanation:_________________________________________

d) Is the result of your test supporting the manufacturer claims that So-Lean brand of hot dogs has an average fat content less than 18g per hot dog? Explain your answer.

Yes No (circle one)

Explanation:___________________________________________

Part two: Multiple choice questions. Select one from A to E as appropriate.

Use following for Questions #6 - #10

Student Study Times.A survey asked the following question: “ About how much time (in minutes) do you study on a typical weeknight?” of randomly selected 46 first-year ASU students. The sample mean was 118 minutes. Let μ be the mean study time of all first-year ASU students, assume that population standard deviation σ = 65 minutes.

Question #6 (7 points) Compute a margin of error in a 90% confidence interval for μ . Give 2 decimal places for the final answer.

A) 19.17 B) 18.78 C) 15.77 D)19.30 E) none of these

Question #7 (7 points)What sample size is needed for a margin of error in 95.44% confidence interval for μto be no more than 5 minutes.

A) n≥650 B) n≥458 C) n≥676 D) n≥277 E) none of these

Question #8 (7 points)Suppose you want to test if μ is less than 2 hours (120 minutes) . Formulate null and alternative hypotheses to be tested:

A) H 0:μ≥118 B) H 0:μ<120 C) H 0:μ=118 D) H 0:μ=120 H a :μ=120 H a :μ=118 H a :μ<118 H a :μ<120 E) none of these

Question#9: (7 points)Suppose the p-value for your test you conducted in previous question (#8) was 0.42, whatis the conclusion for your test at 5% significance level? Select one of the answers below:

(A) Reject H0 , we have no evidence for H a at α=0.05 (B) Do not reject H 0 , we have no evidence for H a at α=0.05 (C) Reject H 0 , we have evidence for H a at α=0.05 (D) Do not reject H 0 , we have evidence for H a at α=0.05

Question#10 (7 points)Suppose you want to test the following hypotheses: H0 : μ=110 versus Ha : μ≠110at 5% significance level. Select critical value(s) for the rejection region for your test:

(A) ±1.645 (B) ±2.014 (C) ±1.28 (D) ±1.96 (E) none of theseUse following information in Questions #11 - #13:

Cholesterol Levels of young males.Suppose a researcher wants to test if a mean cholesterol level ( μ ) of young males who experienced a mild heart attack is higher than that of healthy young males, which isknown to be 188mg/dl. His hypotheses are: H0 : μ = 188 versus Ha : μ > 188

Question#11 (7points)Suppose he knows the population standard deviation and is using a z-test , and the test statistics he received is z = 2.37. Compute the p- value for his test. Give 4 decimal places. A)0.0089 B) 0.9911 C)0.0178 D) 1.9822 (E) none of these

Question#12 (7points)Suppose he does not know the population standard deviation and is using a t-test , his sample size n= 15 and and the test statistics he received is t = 1.90. Estimate the p- valuefor his test from the t-table:

A) 0.025< p−value<0.05 B) 0.05< p−value<0.025 C) 0.05< p−value<0.10 D) 0.01< p−value<0.025 (E) none of these

Question#13 (7points)Suppose null hypothesis was rejected at 5% significance level, only one of the following is the correct conclusion for our hypotheses test, circle the correct answer.

A) We have no evidence at 5% significance level that mean cholesterol level for young males that experienced a mild heart attack is higher than mean cholesterol level for healthy young males

B) We have evidence at 5% significance level that mean cholesterol level for young males that experienced a mild heart attack is higher than mean cholesterol level for healthy young males.

C) We have evidence at 5% significance level that mean cholesterol level for young males that experienced a mild heart attack is lower than mean cholesterol level for healthy young males

D) We have evidence at 5% significance level that mean cholesterol level for young males that experienced a mild heart attack is the same as mean cholesterol level for healthy young males

Question#14 (5 points) (Extra Credit)Suppose the mean annual income for adult women in one city is $28,520 with standard deviation of $5190 and the distribution is left skewed. What is the sampling distribution of the sample mean x̄ for samples of size 49

A) approximately normal distribution B) t distribution with 48 degrees of freedomC) standard normal distribution D) can't specify, because sample is not large enough

FORMULAS

Sampling Distribution of x̄ : μ x̄=μ , σ x̄=σ

√n,

Standardized version of x̄ : z=x̄ −μσ /√n

Studentized version of : x̄ : t=x̄− μs/√n

Confidence Intervals for μ , Confidence level C= (1−α )∗ 100%

Z-interval: x̄ ± zα /2σ

√n Margin of error: E=zα /2

σ

√n

T-interval: x̄ ± t α /2s

√n, df=n-1,

Sample size estimation: n=( zα /2σ

E )2

Hypothesis test for one Population Mean

H 0:μ=μ0 vs H a:μ ≠ μ0 or H a:μ>μ0 or H a:μ<μ0

Z-test ( σ known): test statistics: z=x̄ −μ0

σ /√n

T-test ( σ unknown): test statistics: t=x̄− μ0

s/√n, df=n-1

Key:

Question#1

Distribution is :Approximately Normal, μ x̄ =150 lb σ X̄ =27

√32≈4.77 lb

Question#2

P( X̄>156.25)= 0.0951

z=9156.25-150)/4.77=1.31, area right of 1.31=0.0951

Sketch must show normal curve with center at 150, and shaded area right of 156.25 or N(0,1) with shaded area right of 1.31

Question#3N=36, use T-interval procedure since we have no population SD, but s is given

18.6±2.030( 1

√36 ) gives (18.26, 18.94) df=35 t0.025=2.030

Question#4Select Not CorrectExplanation:Our CI is all above 18 grams, and we have 95% confidence that μ is inside that interval.

Question#5

a) t=17.6−18

1.7

√36

=−1.41

b)Df=35, Critical value= -t0.05= -1.69

Sketch must show t-curve centered at 0, region right of -1.69 is labeled as the nonrejection region, rest as the rejection region

c) Select Do not reject H 0

Explanation: -1.41 falls into nonrejection region

d) Select NOExplanation: H0 is not rejected, so we have no evidence for alternative hypotheses.

Question number #6 #7 #8 #9 #10 #11 #12 #13

Answer: C C D B D A A B

Question number

(extra credit)

#14

Answer: A