Exam 12 july 2013, questions and answers - Semester 1 PDF

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Student ID:Family name:Other names:Desk number: Date:Signature:Examination in the School of Mathematical SciencesSemester 1, 2013005543 STATS 1000 Statistical Practice I102232 STATS 1004 SOLUTIONSTime for completing booklet: 120 mins (plus 10 mins reading time).Question Marks1 /2 /3 /4 /5 /6 /Total ...

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Student ID: Family name: Other names: Desk number:

Date:

Signature:

Examination in the School of Mathematical Sciences Semester 1, 2013 005543

STATS 1000

Statistical Practice I

102232

STATS 1004

SOLUTIONS

Time for completing booklet: 120 mins (plus 10 mins reading time).

Question

Marks

1

/8

2

/8

3

/14

4

/17

5

/14

6

/9

Total

/70

Instructions to candidates • Attempt all questions and write your answers in the space provided below that question. • If there is insufficient space below a question, then use the space to the right of that question, indicating clearly which question you are answering. • Only work written in this question and answer booklet will be marked. • Examination materials must not be removed from the examination room. Materials • Calculators are permitted. • One A4 sheet of hand-written notes (both sides) is permitted • English and foreign-language dictionaries may be used. Do not commence writing until instructed to do so.

Page 1 of 11

Statistical Practice I

8 Total

Page 2 of 11

Question 1. A group of athletes who suffered hamstring injuries were randomly assigned to one of two new exercise programs. In one program athletes engaged in static stretching while in the other program athletes did trunk stabilization exercises. The time taken to return to sports activities was recorded for each athlete. 1(a) Is this a designed experiment or observational study? Briefly justify your answer.

/2 marks

Solution Experiment. Manipulation of treatment by researcher. 1(b) Identify the explanatory and response variables.

/2 marks

Solution Explanatory: Exercise program. Response: Time to return to sports activities. 1(c) Explain why it was important to assign the athletes to the two different treatments randomly.

/2 marks

Solution Allowing athletes to self select could confound results. Other issues such as severity of injury, diet, age, could also affect time to heal. Randomisation should equalise the treatment groups with respect to such variation. is average out the effects of the lurking variables. 1(d) Did this study include a control group?

/2 marks

Solution No. No group that received either no treatment or the standard treatment was included in the study.

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Statistical Practice I

8 Total

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Question 2. 20% of customers of a restaurant only order coffee after the main course. 14% of the customers only order dessert after the main course. 42% order both dessert and coffee. 2(a) What is the probability a randomly chosen customer will order neither coffee nor dessert?

/2 marks

Solution = 1 − P (C ∪ D)

= 1 − [.2 + .14 + .42] = 1 − .76 = .34

2(b) Are the events ordering coffee and ordering dessert independent? Show your reasoning. /2 marks

Solution P (C) = .2 + .42 = .62 P (D) = .14 + .42 = .56 P (C)P (D) = (.62)(.56) 6= P (C ∩ D) = .42 so not independent. 2(c) Are the events ordering coffee and ordering dessert disjoint? Show your reasoning.

/2 marks

Solution Someone can order both coffee and dessert so not disjoint. OR P (C ∩ D) 6= 0 so not disjoint. 2(d) What is the probability that 2 customers, sitting at separate tables, both order only coffee after the main course? Assume that the customers made their choices independently.

/2 marks

Solution P(both order just coffee)=P(1st orders just coffee) × P(2nd

orders just coffee)

= .2 × .2 = .04

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Statistical Practice I

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14 Total Question 3. A lecturer decided to compare the selling prices of textbooks from two internet bookstores. She took a simple random sample of 10 textbooks used that semester in courses at her university. She then recorded the price for each book at both sites. The following summary statistics were recorded. Sample Mean Sample SD

Site A 87.30 34.20

Site B 83.00 31.96

Difference 4.30 4.715

3(a) Explain why this is a paired sample design. /2 marks

Solution There are 10 individuals(books) with 2 variables measured on each individual(price at each store). 3(b) What assumptions are necessary for a paired sample t-test?

/2 marks

Solution Assumes differences are independent N (µD , σD ) observations. 3(c) State the appropriate null and alternative hypothesis.

/2 marks

Solution

H0 : µD = 0 Ha : µD 6= 0

where µD is the population mean of differences.

3(d) Calculate the value of the test statistic and draw a sketch to illustrate the associated p-value. /4 marks

Solution t=

Please turn over for page 5.

¯ d 4.3 √ = √ = 2.88 sd / n 4.715/ 10

Statistical Practice I

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3(e) What is the distribution (including degrees of freedom) of the test statistic if the null hypothesis is true? /2 marks

Solution t(n − 1) = t(9) 3(f) Explain what a Type I error would mean in this context.

/2 marks

Solution The lecturer decides that there is an average price difference, when in fact there isnt.

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Statistical Practice I

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17 Total Question 4. Infants who cry easily may be more easily stimulated than others. This may be a sign of higher IQ. Child development researchers explored the relationship between the crying of infants 4 to 10 days old and their later IQ test scores. 38 infants were selected to participate in the study. A snap of a rubber band on the sole of the foot caused the infant to cry. The researchers measured the intensity of crying by the number of peaks in the most active 20 seconds. They later measured the children’s IQ at age 3 years. Below is the SPSS output from a regression analysis of the data. Use the relevant output to help answer the questions that follow. Table 3.1: SPSS Output for Simple Linear Regression

4(a) State the conditions necessary for inference. Which of the conditions can the plots below be used to assess? Use the plots to assess the

Please turn over for page 7.

Statistical Practice I

Page 7 of 11

validity of those assumptions.

/4 marks

Solution Independence: Normality: of residuals Linearity: The conditional mean µ varies linearly with X. µ = β0 + β1 x. No curved pattern so condition reasonable. Equal standard deviation of Y with respect to x. Valid- roughly equal spread about 0 line from left to right. 4(b) Write down the estimated regression equation. Use the estimated regression equation to predict the IQ of a randomly chosen child with a Cry count of 10.

/3 marks

Solution • The estimated regression equation is c IQ = 91.268 + 1.493 Cry count • 91.268 + 1.493(10) = 106.198

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Statistical Practice I

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4(c) A child in the study with a cry count of 10 had an IQ of 87. Calculate the residual for this child. /3 marks

Solution Residual=observed -expected response= 87 −106.198 = −19.198. 4(d) Determine whether there is a significant linear relationship between IQ and Cry count at the 5% significance level. Give the appropriate null hypothesis, write down the value of the test statistic and the associated p-value. State your conclusion, in context (assuming all assumptions are reasonable).

/4 marks

Solution H0 : β1 = 0 Ha : β1 6= 0 T = 3.065 p-value = .004 Reject H0 at 1% significance level, Evidence of a significant linear relationship between IQ and Cry count. 4(e) Below is part of the SPSS data editor. Would it be unusual for a child with a crying rate of 20 to have an IQ of 80? Explain your answer.

/2 marks

Solution Yes. The value is outside the 95% prediction interval. 4(f) Do you believe that using crying rate alone to predict IQ will give an accurate prediction? Refer to the relevant information from the analysis to support your answer.

/1 mark

Solution No. R2 is quite small. Also the prediction intervals are large in comparison to the range of observed IQs.

Please turn over for page 9.

Statistical Practice I

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14 Total Question 5. 72 anorexic teenage girls were randomly assigned to one of three psychological treatments.The three therapy regimes were cognitive therapy, family therapy and no therapy. The change in weights (in pounds) was recorded at the end of the study for each girl. Analysis of variance was used to compare the weight changes. Below is the summary information for the data.

Table 4.1: Descriptive Summary

5(a) State the assumptions required for the analysis. /2 marks

Solution • The observations are independent both within and between groups. • The observations are Normally distributed within each group. • The standard deviation is the same for all groups.

5(b) Write down null and alternative hypotheses for the AVOVA defining any parameters you use. /3 marks

Solution H0 : µ1 = µ2 = µ3

Ha : not all µi are equal

where µ1 is the mean weight change for girls who have no therapy µ2 is the mean weight change for girls who have cognitive therapy µ3 is the mean weight change for girls who have family therapy

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Statistical Practice I

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5(c) What is the distribution of the test statistic for the ANOVA, including degrees of freedom, if H0 is true? /2 marks

Solution F (2, 69) 5(d) Using the ANOVA table, calculate tyhe value of the test statistic and write down the p-value for this test. Explain why you reject or retain the null hypothesis at the 5% significance level.

/4 marks

Solution f = 307 .22/56.677 = 5.421

p-value= .006. Reject H0 at

5% significance level as p=value< .05.

5(e) Below is the pairwise comparisons of means output. Can you conclude that cognitive therapy is more effective than having no treatment at all? Justify your conclusion with reference to the relevant hypothesis test.

/3 marks

Solution H0 : µ1 = µ2 has a associated p-value of .212. No evidence of a mean difference in weight change between the two therapies.

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Statistical Practice I

9 Total

Page 11 of 11

Question 6. Soldiers at Fort Gordon, Georgia and Fort Campbell, Kentucky completed a questionnaire, which included items about cigarette use, alcohol consumption, and coffee consumption (Zvela, Barnett, Smedi, Istvan, & Matarazzo, 1990). One of the questions the researchers wanted to answer was the following: Is there a relationship between smoking and gender in the military? The data are below.

6(a) State null and alternative hypotheses for the chi-square test of independence on these data. /2 marks

Solution H0 : Gender and smoking status are independent Ha : Gender and smoking status are not independent 6(b) Calculate the expected number of female ex-smokers under H0 .

/2 marks

Solution =

91(126) 610

= 18.8

6(c) Write down the formula for the test statistic, and its distribution under the null hypothesis. /3 marks

Solution X 2 =

P (obs−exp)2 exp

2 = χ2(3−1)(2−1) = χ22 under H0 . 6(d) has a χ(r−1)(c−1)

From the output, explain whether the null hypothesis should be retained or rejected at the 1% significance level. /2 marks

Solution X 2 = 12.724 has p-value= .002 < .01 so reject H0 at the 1% significance level.

End of examination questions....