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STAT 350

by: Joshua Robinson

STAT 350 STAT 350

Joshua Robinson
GPA 3.5

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These are practice questions for STAT 350, Exam II. They go over Critical values, hypotheses, proportions, test statistics, etc.
Business Economic Statistics 1
Karen Keating
Study Guide
Math Statistics Exam Questions
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This 7 page Study Guide was uploaded by Joshua Robinson on Monday April 11, 2016. The Study Guide belongs to STAT 350 at Kansas State University taught by Karen Keating in Spring 2016. Since its upload, it has received 159 views. For similar materials see Business Economic Statistics 1 in Statistics at Kansas State University.


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Date Created: 04/11/16
1. A population has a standard deviation of 12. A random sample of 36 observations taken  from this population produces a standard deviation of 10.9. Which of these Numbers are  parameters and which are statistics? a. 12 is a parameter and 10.9 is a statistic b. 12 is a statistic and 10.9 is a parameter c. Both 12 and 10.9 are parameters d. Both 12 and 10.9 are statistics Answer: A The standard deviation of 12 is from a population which is why it is a parameter, and the  standard deviation of 10.9 is from the sample, making it a statistic.  2. From a population that is normally distributed, a sample of 25 elements is selected and  the mean and standard deviation of the sample is computed. For the interval estimation of u, the proper distribution is the  a. Normal Distribution b. T distribution with 25 degrees of freedom  c. T distribution with 26 degrees of freedom d. T distribution with 24 degrees of freedom  Answer: D Look for the population standard deviation (SD). If the population SD is given you use  normal distribution. Since it is a SD from a sample, you use t distribution and (n­1) from the  sample number given.  3. The average hourly wage of computer programmers with 2 years of experience has been  21.80. Because of high demand for computer programmers, it is believed there has been a significant increase in the average wage of computer programmers. To test whether or not there has been an increase, what are the correct hypotheses? a. H(o) u < 21.80        H(a) u > 21.80 b. H(o) u = 21.80        H(a) u (does not equal) 21.80 c. H(o) u > 21.80        H(a) u < 21.80 d. H(o) u < 21.80        H(a) > 21.80 Answer: D  The null hypothesis comes from the first statement, stating that the average hourly wage  has been 21.80 for a computer programmer (H(o) u < 21.80). The alternative hypothesis comes  from the statement that the average wage has been believed to be increased (H(a) u > 21.80).  4. The values for a p­value must be between  a. Zero and infinity b. Minus infinity and plus infinity c. Zero and one d. ­1 and +1 Answer: C The p­value is represented by positive decimals between zero and one 5. “D” size batteries produed by MNM Corporation has had a life expectancy of 87 hours.  Because of improved production process, it is believed that there has been an increase in  the life expectancy of its “D” size batteries. A sample of 36 batteries showed an average  life of 88.5 hours. Assume from past information that it is known that the standard  deviation of the population is 9 hours. We want to test the hypotheses H(o) u < 87 H(a) u > 87 Compute the p­value  a. 0.1587 b. 0.0794 c. 0.3174 d. None of these are the correct p­value Answer: A TS = (88.5­87/(9/sqrt36) = 0.8413 1 – 0.8413 = 0.1587 6. A carpet company advertises that it will deliver your carpet within 15 days of purchase. A sample of 9 past customers is taken. IN the sample, the average delivery time was 16.2  days with standard deviation 5.6 days. Construct a 95% confidence interval for the true  mean delivery time for all deliveries made by this company. a. 10.7 to 19.3 days b. 11.9 to 20.5 days c. 12.5 to 19.9 days d. None of these are correct Answer: C (check answer, on paper it says B) 1.96*(5.6/sqrt9) = 3.66 16.2 ­3.66 = 12.5, 16.2 + 3.66 = 19.9 7. A simple random sample of 64 observations was taken for a large population. The sample mean and the standard deviation were determined to be 320 and 120 respectively. What is the standard error of the mean? a. 1.875 b. 40 c. 5 d. 15 Answer: D 120/sqrt64 = 15 8. The average price of homes sold in the U.S in 2012 was $240,000. In the same year, a  sample of 144 homes sold in Chattanooga showed an average price of $246,000, with  standard deviation $36,000. We are interested in determining whether or not the average  price of all homes sold in Chattanooga is more than the national average, so we want to  test the hypotheses: H(o) u < 240,000 H(a) u > 240,000 The P­value is 0.0237. At significance level a = 0.01, what is the conclusion of the test? a. Reject H(o) and conclude that the average prince in Chattanooga is higher than  the national average. b. Reject H(o)  and conclude that the average price in Chattanooga is NOT higher  than the national average. c. Do NOT reject H(o) and conclude that the average price in Chattanooga is higher  than the national average. d. Do NOT reject H(o) and conclude that the average price in Chattanooga is NOT  higher than the national average  Answer: D Since the p­value = 0.0237 > a = 0.01 you would NOT reject H(o) and conclude that the average  price in Chattanooga is NOT higher than the national average.  9. When the p­value is used for hypothesis testing, we reject the null hypothesis if  a. P­value < a b. A < p­value c. P­value d. P­value < 1 – a  Answer: A Because the p­value is less than a, we reject the null hypothesis.  10. The academic planner of a university thinks that less than 35% of the entire student body  attends summer school. The correct set of hypotheses to test his belief is a. H(o): p > 0.35 vs. H(a): p > 0.35 b. H(o): p < 0.35 vs. H(a): p > 0.35 c. H(o): p > 0.35 vs. H(a): p < 0.35 d. H(o): p  >0.35 vs. H(a): p < 0.35 Answer: C H(o): p > 0.35 vs. H(a): p < 0.35 The academic planers statement is the alternative hypothesis in this situation. Therefore  H(a): p < 0.35 11.  A sample of 225 elements from a population with a standard deviation of 75 is selected.   The sample mean is 180. The 95% confidence interval for u is a. 105 to 225 b. 175 to 185 c. 100 to 200 d. 170.2 to 189.8 Answer: D s (sample standard deviation) = 75     Xbar (sample mean) = 180   n (sample quantity) = 225 MOE = Z(a/2)*s/sqrt(n) MOE = 1.96*(75/sqrt225) = 9.8 180 – 9.8 = 170.2, 180 + 9.8 = 189.8 12. As a rule of thumb, the sampling distribution of the sample proportion can be  approximated by a normal probability distribution whenever a. np > 5 b. np > and n(1­p) >5 c. n > 30 d. None of these alternatives is correct Answer: B In order to use normal probability distribution, np > 5 and n(1­p) > 5 13. In determining the sample size necessary to estimate a population proportion, which of  the following information is NOT needed? a. The maximum margin of error that can be tolerated b. The confidence level required c. A preliminary estimate of the true population proportion p d. The mean of the population Answer: D The equation to determine the sample size is, n = (Za/2)^2)*p(1­p)/E^2 which does not  include the mean of the population.  14. A population has a mean of 300 and a standard deviation of 18. A sample of 144  observations will be taken. What is the probability that the sample mean will be between  299 and 301? a. 0.4972 b. 0.8664 c. 0.0638 d. 0.9756 Answer: A P(299 < x(bar) < 301)  p((299­300)/(18/sqrt144)) <  Z  < ((301­300)/(18/sqrt144)) P(­0.67 < Z < 0.67)  (numbers from standard distribution table using ­0.67 and 0.67)  0.7486 – 0.2514 = 0.4972 15.  In order to estimate the average electric usage per month, a sample of 81 houses was  selected, and the electric usage was determined. Assume a population standard deviation  of 450 kilowatt hours. At 95% confidence, what is the margin of error? a. 1.96 b. 50 c. 98 d. 42 Answer: C  MOE = Confidence Interval*(pop. Standard deviation/sqrt(sample population)   MOE = 1.96*(450/sqrt81) = 98 16. A sample size 36 is taken form a normal population. The sample mean is 24.6 and sample standard deviation is 12. We want to use this information to conduct the hypothesis test H(o): u < 20 H(a): u > 20 After some calculations, we determine the p­value is 0.0137. If the hypotheses are  changed to a two­sided test, what is the p­value? a. 0.0137 b. 0.0274 c. 0.00685 d. None of these alternatives are correct. Answer: B  Because it is a two sided test, double the p­value. = 0.0274 17. The building specifications in a certain city require that the sewer pipe used in residential  areas have a mean breaking strength of at least 2500 pounds per lineal foot. A  manufacturer would like to supply the city with sewer pipe, but the city wants to make  sure the pipe meets specifications. The city will collect a sample of pipes form the  manufacturer in order to perform the following hypothesis test. (u is the mean breaking  strength of all pipes made by this manufacturer) H(o) u > 2500 (manufacturer’s pipe meets specifications) H(a) u < 2500 (manufacturers pipe does not meet specificatinos) In the context of this problem, what is a Type I error? a. The pipe does not meet specifications, and the city concludes that it does. b. The pipe does not meet specifications, and the city concludes that it does not. c. The pipe meets specifications, and the city concludes it does not. d. The pipe meets specifications, and the city concludes that it does not. Answer: D Type I error = Reject H(o) when it is true.  18. A random sample of 64 students at a university showed an average age of 25 years and a  sample standard deviation of 2 years. We want to construct a 98% confidence interval for the true average age of all students in the university. What is the critical value? a. 1.96 b. 1.645 c. 2.387 d. 2.58 Answer: C a = 0.02 19.  Four hundred people were asked whether gun laws should be more stringent. Three  hundred said “yes” and 100 said “no.” Construct a 99% confidence interval for the  proportion in the population who will respond “no”. a. 0.194 to 0.306 b. 0.208 to 0.292 c. 0.694 to 0.806 d. 0.247 to 0.253 Answer: A P(bar) = 100/400 = 0.25 1 – a = .99  a = 0.1  .01/2  0.005  Because this is a proportion we will use normal distribution. a/2 = 0.005, so the area to left is . 995, which we then take to the normal distribution table to come up with Critical Value = 2.575 Standard Error (SE) = sqrt(p(bar)*(1­p(bar)/n)  sqrt(0.25*0.75)/400) = 0.02165 MOE = (SE)(CV) = .02165*2.575 = 0.056  .25 ­ .056 = 0.194, and .25 + .056 = 0.306 20. From a population of cans of coffee marked “12oz”, a sample of 50 canns was selected  and the contents of each can were weighted. The sample revealed a mean of 11.8oz with  a standard deviation of 0.5oz. We want to test the hypotheses H(o): u = 12 H(a): u =(does note equal) 12 Compute the test statistic a. 2.83 b. ­2.83 c. 2.00 d. ­2.00 Answer: B T.S = (x(bar) – u(o)/(s/sqrt(n)) = (11.8 – 12)/(0.5/sqrt50) = ­2.83


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