Exhibit 101 Salary information regarding male and female employees of a large company is shown below.
Male Female Sample Size 64 36 Sample Mean Salary (in $1,000) 44 41 Population Variance (σ²) 128 72
Solution: margin of error is: z(0.025)√(128/45+72/36) = 1.96(2) = 3.92
Refer to Exhibit 101. At 95% confidence, the margin of error is a. 1.645
b. 1.96
c. 2.000
d. 3.920 _____________ 2. Refer to Exhibit 101. If you are interested in testing whether or not the average salary of males is significantly greater than that of females, the test statistic is
Solution: statistic = (4441)/ √(128/45+72/36) = 3/2 = 1.5
1.96
b. 2.0
c. 1.5
d. 1.645
________________________________ 4. Exhibit 103
Today Five Years Ago x̄(xbar) 82 88
σ² 112.5 54
n 45 36 Refer to Exhibit 103. The pvalue for the difference between the two population means is
Solution: statistic = (8288)/√112.5/45+54/36) = 6/2 = 3 pvalue = 2P(z<3) = 2(0.0013) = 0.0026
a. .0026 b. .4987 c. .9987 d. .0013 _____________________________________________________ 5.
Exhibit 121 When individuals in a sample of 150 were asked whether or not they supported capital punishment, the following information was obtained.
Do you support capital punishment? # of individuals Yes 40 No 60 No Opinion 50
We are interested in determining whether or not the opinions of the individuals (as to Yes, No, and No opinion) are uniformly distributed.
The conclusion of the test (at 95% confidence) is that the
Solution:
Ho: Opinions are uniformly distributed Ha: Opinions are not uniformly distributed
Yes No No opinión Total Fo 40 60 50 150 Fe 50 50 50 150 (FoFe)^{2}/Fe 2 2 0 4
Statistical value = 4 Df = 2, pvalue = P(chi^{2} > 4) = 0.1353 > 0.05
We fail to reject Ho
a. test is inconclusive
b. distribution is not uniform
c. distribution is uniform
d. none of these alternatives is correct
_______________________________
6. Exhibit 122 Last school year, the student body of a local university consisted of 30% freshmen, 24% sophomores, 26% juniors, and 20% seniors. A sample of 300 students taken from this year’s student body showed the following number of students in each classification Freshmen 83 Sophomores 68 Juniors 85 Seniors 64
We are interested in determining whether or not there has been a significant change in the classifications between the last school year and this school year.
Solution:
Ho: Opinions are uniformly distributed Ha: Opinions are not uniformly distributed
Freshmen Sophomores Juniors Seniors Total Total Fo 83 68 85 64 300 Fe 90 72 78 60 300 (FoFe)^{2}/Fe 0.5444 0.2222 0.6282 0.2667 1.6615
Statistical value = 1.6615
Refer to Exhibit 122. The calculated value for the test statistic equals a. 300
b. 1.6615
c. 6.6615
d. 0.5444
_________________________________
7. In a sample of 100 Republicans, 60 favored the President's antidrug program. While in a sample of 150 Democrats, 84 favored his program. At 95% confidence, test to see if there is a significant difference in the proportions of the Democrats and the Republicans who favored the President's antidrug program.
a.State the hypotheses involved in this test. Η 0: P1  P2  Select your answer greater thangreater than or equal to equal to less than or equal toless thannot equal toItem 1 0 Η a: P1  P2  Select your answer greater thangreater than or equal toequal toless than or equal toless than not equal to Item 2 0
b.Compute the value of the z test statistic (to 2 decimals).
Solution: p1hat = 60/100 = 0.6 p2hat = 84/150 = 0.56 phat = (60+84)/(100+150) = 0.576 Statistic = z = (0.60.56)/√[0.56(10.56)(1/100+1/150)] = 0.6269623 Answer: 0.63
c.What is the pvalue (to 4 decimals)? Solution: pvalue = 2P(z>0.63) = 0.5287 Answer: 0.5287
d.What is your conclusion?  Select your answer Conclude there is a significant difference between the proportions of Democrats and Republicans who favor the president's antidrug plan Do not conclude there is a significant difference between the proportions of Democrats and Republicans who favor the president's antidrug plan
_______________________________________________
8.
For a onetailed test (upper tail), a sample size of 26 at 90% confidence, t = a. 1.316
b. 1.740
c. 1.740
d. 1.316
__________________________________________________
9. 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, the correct hypotheses to be tested are a. H0: μ > 21.80 Ha: μ ≤ 21.80
b. H0: μ = 21.80 Ha: μ ≡ 21.80
c. H0: μ ≤ 21.80 Ha: μ > 21.80
d. H_{0}: μ < 21.80 H_{a}: μ ≥ 21.80
10. A weatherman stated that the average temperature during July in Chattanooga is 80 degrees or less. A sample of 32 Julys is taken. The correct set of hypotheses is  a.  H_{0}: μ ≤ 80 H_{a}: μ > 80 
  b.  H_{0}: μ ≡ 80 H_{a}: μ = 80 
  c.  H_{0}: μ ≥ 80 H_{a}: μ < 80 
  d.  H_{0}: μ < 80 H_{a}: μ > 80 

11. A group of young businesswomen wish to open a high fashion boutique in a vacant store but only if the average income of households in the area is at least $25,000. A random sample of 9 households showed the following results. $28,000 $24,000 $26,000 $25,000 $23,000 $27,000 $26,000 $22,000 $24,000
Assume the population of incomes is normally distributed. 1. Compute the sample mean and the standard deviation (to 2 decimals, if necessary). x ¯ = 25,000 and σ = 1936.49 2. State the hypotheses for this problem. Η _{0}: μ  Select your answer greater than greater than or equal to equal toless than or equal toless thannot equal to 25,000 Η _{a}: μ  Select your answer greater thangreater than or equal toequal toless than or equal to less than not equal to 25,000 3. Compute the test statistic. Answer: 0 12. The average starting salary of students who graduated from colleges of Business in 2009 was $48,400. A sample of 100 graduates of 2010 showed an average starting salary of $50,000. Assume the standard deviation of the population is known to be $8,000. We want to determine whether or not there has been a significant increase in the starting salaries. 1. State the null and alternative hypotheses to be tested. H_{o}: μ ≤ 48,400   H_{a}: μ > 48,400   2. Compute the test statistic. Answer: 2 3. The null hypothesis is to be tested at 95% confidence. Determine the critical value for this test (to 2 decimals). Answer: 1.65 We reject Ho (there has been a significant increase in the starting salaries)
5. Compute the pvalue (to 4 decimals). % Answer: 0.0228 13. A department store believes that telephone calls come into the switchboard at 10minute intervals, according to a Poisson distribution. Before ordering new equipment, the store wishes to determine whether the Poisson model is a valid assumption. Records on the number of calls received were kept for a random selection of 150 tenminute intervals. The results are shown below. Number of Calls Frequency 0 5 1 18 2 24 3 30 4 32 5 13 6 20 7 8 ___ 150 1. What is the average number of calls during these tenminute intervals (to 1 decimal)? Answer: 3.5 2. Generate the expected number of calls using a Poisson probability table (to 3 decimals). Number of Calls  e_{i}  0 or 1  20.383  2  27.744  3  32.368  4  28.322  5  19.825  6  11.565  7 or more  9.793  3. 4. Give the null and alternative hypotheses for the appropriate test. Η _{0}:  Select your answer The number of telephone calls during a 10 minute interval follows a Poisson distributionThe number of telephone calls during a 10 minute interval does not follow a Poisson distribution Η _{a}:  Select your answer The number of telephone calls during a 10 minute interval follows a Poisson distributionThe number of telephone calls during a 10 minute interval does not follow a Poisson distribution 5. Determine the number of degrees of freedom for this test. Let k = number of categories or classes remaining after combining classes k = 7, Df = k1 = 71 = 6 Answer: df = 6 6. Calculate the value of the test statistic (to 2 decimals). χ^{2} = (2320.343)^{2}/20.343 +…..+(89.793)^{2}/9.793 = 9.99 Answer: 9.99 7. Based on the pvalue what is your conclusion? If we use = 0.05 pvalue = P(chi>9.99)= 0.1251 > 0.05 Answer: we fail to reject Ho  Select your answer The Poisson distribution is not a valid model The Poisson distribution is a valid model
