Exam 3 Solution

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STAT 34500/51100 Statistics Exam 3 Review Part 1- Solution

Q.N. 1) Suppose a random variable X has a mean of 20 and a standard deviation of 5. Calculate the mean and the standard deviation of sample mean for each of the following sample size a) n = 40 b) n = 75 Solution: We have µ = 20 and σ = 5 Hencce ¯ = µ = 20 and σX¯ = a) For n = 40, E(X) ¯ = µ = 20 and σX¯ = b) For n = 75, E(X)

√σ n √σ n

= =

√5 40 √5 75

= 0.791 = 0.577

Q. N. 2) Suppose there is a normally distributed population with mean 70 and variance 64. If X be the average of a sample of 50, find the following probabilities a) P (X ≤ 75) b) P (68 ≤ X ≤ 72) Solution: We have µ = 70 and σ 2 = 64 and n = 50 a) ( ) ( ) 75 − 70 √ P X ≤ 75 = P Z ≤ = P (Z ≤ 4.419) ≈ 1 8/ 50 b)

( P (68 ≤ X ≤ 72) = P

68 − 70 72 − 70 √ √ ≤Z≤ 8/ 50 8/ 50

) = 0.9232

Q.N. 3) If a sample of size 50 is drawn from a population that has mean 27 and standard deviation 9, what is the probability that the mean of the sample will be less than 28? ¯ ≤ 28) Solution: We have n = 50, µ = 27, σ = 9 We want to calculate P (X We have ) ( 28 − 27 ¯ √ P (X ≤ 28) = P Z ≤ = P (Z ≤ 0.79) = 0.7852 9/ 50 Q.N. 4) Let X1 , X2 , X3 , X4 , X5 be a random sample from a binomial distribution with n = 20 and p unknown. a) Show that X/20 is an unbiased estimator of p. b) Estimate p based on the data: 3, 4, 4, 5, 6. Solution: a) We know that ( ) E X/20 =

1 E(X) 20 1

1 µ since E(X) = µ 20 1 = np sinceE(X) = µ = np 20 1 = 20 × p 20 = p =

Therefore, X/20 is an unbiased estimator of p. b) Since X/20 is an unbiased estimator of p we have pˆ =

X 4.4 = = 0.22 20 20

Q.N. 5) Let X1 , X2 , · · · , Xn be a sample from an exponential distribution with pdf f (x) = 1θ exp(−x/θ). Find the method of moment and maximum likelihood estimator of θ Solution: ∫



E(X) =

xf (x)dx ∫

0



= ∫

0

1 = θ = θ ⇒ E(X) = θ

1 x exp(−x/θ)dx θ ∞

x exp(−x/θ)dx 0

Note that we have only one parameter to estimate. Therefore we have X = E(X) = θ. This implies θˆ = X. To obtain the maximum likelihood estimator we have the likelihood function given by

L(θ) =

n ∏

f (θ, xi )

i=1 n ∏

1 exp(−xi /θ) θ i=1 ( ) ( )n n ∑ 1 1 = exp − xi θ θ i=1 =

Now the log-likelihood function is given by L = log L 2

= log

( n ∏ (

) f (θ, xi )

i=1

= log θ

−n

(

1∑ exp − xi θ i=1 n

))

1∑ xi θ i=1 n

= −n log(θ) − ⇒ Now setting Therefore,

dL dθ

dL n 1 ∑ =− + 2 xi dθ θ θ ∑

= 0 we get θ =

xi n

= X. θˆ = X.

Q.N. 6) Find zα/2 for the following confidence level. a) 95% b) 88% c) 85%. Answer: a) 1.96 b) 1.555 c) 1.44 Q.N. 7) Construct a 90% confidence interval for the mean of a normal population with a standard deviation of 10 if a random sample of size 80 from the population has sample mean 45. Solution: We have σ = 10, n = 80, X = 45. Since we want to construct 90% confidence interval, α = 0.1. We know that (1 − α) confidence interval for the population mean is given by σ X ± zα/2 √ n Since α = 0.1, α/2 = 0.05, therefore Zα/2 = 1.645 Hence the 90% confidence interval for µ is 10 45 ± 1.645 √ 80 which yields, the required interval (43.16, 46.84). Q.N. 8) A random sample, consisting of the values listed below, was taken from a population which is normally distributed. Assuming that the population variance is unknown construct a 95% confidence interval for population mean. 12, 13, 14, 15, 15, 16, 17, 18, 20, 25, 27, 28, 29, 20, 25 3

Solution: We have for the given data the sample mean X = 19.6, the sample variance S 2 = 33.54 therefore s = 5.79. We would like to compute 95% confidence interval for the population mean µ. Since the population variance is unknown the 1 − α confidence interval is given as s X ± tα/2,n−1 √ n We have n=15, therefore the required confidence interval will be 5.79 19.6 ± t0.025,14 √ 15 which yield the required 95% confidence interval as (16.4, 22.8). Q.N. 9) A technician would like to find the sample size for a study of the average time required to fuel an engine. Based on the past experience it is believed that the data will have standard deviation 6 minutes. To be 90% confident what sample size will be needed with an accuracy of 1 minute. Solution: We have σ = 6, α = 0.1 and Error = 1. We know that the minimum required sample size is given by ( ) Zα/2 σ 2 n≥ Error We have ( ) ( )2 Zα/2 σ 2 1.645 × 6 = = 97.4169 Error 1 Therefore, the minimum sample size n = 98. Q.N. 10) Determine the critical value(s) of the test statistic for each of the following large sample tests for the population mean. a) Left tailed test, α = 0.01 b) Right tailed test, α = 0.1. c) Two tailed test, α = 0.05 Answer: a) z = −2.325 b) z = 1.28 c) z = 1.96& − 1.96 Q.N. 11) A random sample of 200 observations indicated a mean of 4,117 with a standard deviation of 300. a) Set up the null and alternative hypotheses to test the mean is greater than 4,100. b) If α = 0.01 find the critical value for the test c) Compute the value of test statistic z and test the hypothesis. Solution: a) H0 : µ ≤ 4100,

Ha : µ > 4100 4

b) Critical value =2.325 X−µ √ = 0.8014, c) Test statistic Z = σ/ n Decision: Fail to reject the null hypothesis as 0.8014 < 2.325 Q.N. 12) A random sample of 49 observations from a population with variance σ 2 = 9 produces a sample mean of 16. Test the hypothesis that the mean is not equal to 15 at α = 0.05? Solution: We would like to test the following hypothesis H0 : µ = 15 Ha : µ ̸= 15 ¯ = 16 We have n = 49, σ = 3, X So the test statistic is

¯ −µ X 16 − 15 √ √ = = 2.33 σ/ n 3/ 49 Note that it is a two tailed test and the critical value(s) is given as Zα/2 = Z0.025 = ±1.96 Decision: Reject the null hypothesis as 2.33 > 1.96. Z=

Q.N. 13) Determine the critical values of the test statistic for each of the following small sample tests for the population mean where the assumption of normality is satisfied. a) left tailed test, α = 0.1 and n = 15 b) right tailed test, α = .1 and α = 20 c) two tailed test, α = 0.05 and n = 8 Answer: a) -1.345 b) 1.328 c) t = 2.365 and −2.365 Q.N. 14) The admitting office at Sisters of Mercy Hospital wants to be able to inform patients of the average level of expenses they can expect per day. Historically, the average has been approximately $ 1,240. The office would like to know if there is evidence of an increase in the average daily billing. 20 randomly selected patients have an average daily charge of $ 1,419. with a standard deviation of $ 342. a) What is the population being studied? b) Conduct the hypothesis test to determine whether there is evidence that average daily charges have increased at α = 0.1 c) What assumption did you make in performing the test in part b? Solution: a) The population is the patients of the Sisters of Mercy Hospital b) H0 : µ ≤ 1240, Ha : µ > 1240 Critical value =1.328 and tα = 3.28 hence, the decision is reject the null hypothesis 5

c) We assumed that the daily charges for patients of Sisters of Mercy Hospital have an approximately normal distribution Q.N. 15) If X1 , X2 , ...., X10 are iid N (0, 1) variables. Find the following probabilities (a) P (X1 + 2X2 + 3X3 ≥ 3) and 2 (b) P (X22 + X42 + X62 + X82 + X10 ≥ 9.24). Solution: We know that X1 , X2 , ....X10 are iid N (0, 1) So Y = X1 + 2X2 + 3X3 has normal distribution with mean 0 and variance 14. We want to find P (Y ≥ 3) where Y ∼ N (0, 14). We have Y −0 3−0 3 P (Y ≥ 3) = P ( √ ≥ √ ) = P (Z ≥ √ ) = 1 − P (Z ≤ 0.802) = 1 − 0.7939 = 0.2061 14 14 14 2 In order to find P (X22 + X42 + X62 + X82 + X10 ≥ 9.24) note that Xi2 ∼ χ2 (1) for i = 1, 2, 3..., 10 2 Therefore Y = X22 + X42 + X62 + X82 + X10 has χ2 distribution with 5 degrees of freedom. We want to find P (Y ≥ 9.24) where Y ∼ χ2 (5) From the χ2 table we have P (Y ≥ 9.24) = 0.1

Q.N. 16) A random sample of 27 auto claims filed with an insurance company produced a mean amount for these claim equal to $2100 with a standard deviation of $460. Construct a 95% confidence interval for the corresponding population mean. Assume that the amounts of all such claims filed with this company have a normal distribution. Solution: We have n = 27, X = 2100, s = 460, α = 0.05. We know that (1 − α)100% confidence interval is given as s X ± tα/2,n−1 √ n Hence, required confidence interval will be 460 2100 ± t0.025,26 √ 26 which yield the required 95% confidence interval as (1914.52, 2285.48). Q.N. 17) Researchers studying the effects of diet on growth would like to know if a vegetarian diet affects the height of a child. The researcher randomly selects 12 vegetarian children that are six years old. The average height of the children is 42.5 inches with a standard deviation of 3.8 inches. The average height of for all six year old children is 45.75 inches. Conduct a hypothesis test to determine whether there is overwhelming evidence at α = 0.05 that six year old vegetarian children are not the same height as the other six year old children. Solution: We would like to test the following hypothesis H0 : µ = 45.75 Ha : µ ̸= 45.75 6

We have n = 12, X = 42.5, s = 3.8, α = 0.05 The test statistic is ¯ −µ X 42.5 − 45.75 √ t= √ = = −2.96 s/ n 3.8/ 12 Note that we have a small sample two tailed test hence the critical values are given as tα/2,n−1 = t0.025,11 = ±2.201 Since the observed test statistic -2.96 is less than -2.201,we reject the null hypothesis concluding the height of that six year old vegetarian children are not the same height as the other six year old children. Q.N. 18) Suppose we have a random sample of size 25 from a normal population with an unknown mean µ and a standard deviation of 4. We wish to test the hypothesis H0 : µ = 10 vs H1 : µ > 10 Let the rejection region be defined by, reject H0 if the sample mean X > 11.2. a) Find α. (b) Find β for H1 : µ = 12. Solution: We know that α = P (Type I Error) = P (Reject H0 |H0 is true) = P (X ≥ 11.2|µ = 10) ( ) X − µ0 11.2 − 10 √ ≥ √ = σ/ n 4/ 25 = P (Z ≥ 1.5) ≈ 0.0668. and β = P (Type II Error) = P (Fail to Reject H0 |H1 is true) = P (X ≤ 11.2|µ = 12) ( ) X − µ1 11.2 − 12 √ ≤ √ = σ/ n 4/ 25 = P (Z ≤ −1.0) = 0.1587 Therefore the power of this test Power = 1 − β = 1 − 0.1587 = 0.8413

7

Q.N. 19) It is claimed that sport car owners drive on the average 18,000 miles per year. A consumer firm feels that the average milage is probably lower. To check, the consumer firm obtained information from 40 randomly selected sport car owners that resulted in a sample mean of 17,463 miles with a sample standard deviation of 1,348 miles. What can we conclude about this claim? Use α = 0.01 Solution: We would like to test the following hypothesis H0 : µ ≥ 18000 Ha : µ < 18000 We have n = 40, X = 17463, s = 1348, α = 0.01 The test statistic is ¯ −µ X 17463 − 18000 √ = −2.52 Z= √ = s/ n 1348/ 40 Note that we have a large sample left tailed test hence the critical value is given as Z0.01 = −2.325 Since the observed test statistic -2.52 is less than -2.325,we reject the null hypothesis. Hence there is not enough evidence to claim that the sport car owners drive on the average 18,000 miles per year. Q.N. 20) In a frequently traveled stretch of I-94 highway, where the posted speed is 70 mph, it is thought that people travel on the average of at least 75 mph. To check this claim, the following radar measurements of the speeds (in mph) is obtained for 10 vehicles traveling on this stretch of the interstate highway. 66, 74, 79, 80, 69, 77, 78, 65, 79, 81 Do the data provide sufficient evidence to indicate that the mean speed at which people travel on this stretch of highway is at least 75 mph? Test the appropriate hypothesis using α = 0.01. Solution: We would like to test the following hypothesis H0 : µ ≥ 75 Ha : µ < 75 We have n = 10, X = 74.8, s = 5.996, α = 0.01 The test statistic is ¯ −µ X 74.58 − 75 √ = −0.22 t= √ = s/ n 5.996/ 10 Note that we have a small sample left tailed test hence the critical value is given as t0.01,9 = −2.821 Since the observed test statistic -0.22 is greater than -2.821,we fail to reject the null hypothesis. Hence there is sufficient evidence to indicate that the mean speed at which people travel on this stretch of highway is at least 75 mph.

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