Complete the following task:
Task A
The manager of a grocery store claims that the average time that customers spend in checkout lines is 20 minutes or less. A sample of 36 customers is taken. The average time spent on checkout lines for the sample is 24.6 minutes; and the sample standard deviation is 12 minutes. Conduct a hypothesis test (at 0.05 level of significance) to determine if the mean waiting time for the customer population is significantly more than 20 minutes. ANSWER QUESTIONS 1 - 4.
1. The correct hypotheses for the test are:
a. H0 is: μ ≥ 20; Ha is: μ < 20
b. H0 is: μ ≤ 20; Ha is: μ > 20
c. H0 is: μ > 20; Ha is: μ ≤ 20
d. H0 is: μ < 20; Ha is: μ ≥ 20
2. The observed value of the test statistic is:
a. 2.3 b. 0.38 c. -2.3 d. -0.38
3. The p-value is:
a. 0.5107 b. 0.0214 c. 0.0137 d. 0.4893
4. The null hypothesis should:
a. not be rejected
b. be rejected
c. the test is inconclusive
d. none of the above answers are correct
Task B
A random sample of 16 students selected from the student body of a large university had an average age of 25 years with a standard deviation of 2 years. Conduct a hypothesis test (at 0.05 level of significance) to determine if the average age of the population of students at the university is significantly different from 24 years. Assume that the distribution of the population of ages is normal.
BASED ON THIS TASK- ANSWER QUESTIONS 5 THROUGH 7 BELOW.
5. The correct hypotheses for the test are:
a. H0 is: μ ≥ 24; Ha is: μ < 24
b. H0 is: μ ≤ 24; Ha is: μ > 24
c. H0 is: μ is not equal to 24; Ha is: μ = 24
d. H0 is: μ = 24; Ha is: μ is not equal to 24.
6. The observed value of the test statistic is:
a. 1.96 b. 2.00 c. 1.645 d. 0.05
7. It can be concluded that the mean age of the student population is:
a. not significantly different from 24
b. significantly different from 24
c. significantly less than 24
d. significantly less than 25
Task C: Use the following to answer questions 8-9:
A major airline company is concerned that its proportion of late arrivals has substantially increased in the past month. Historical data shows that on the average 18% of the company airplanes have arrived late. In a random sample of 1,240 airplanes, 310 airplanes have arrived late. If we are conducting a hypothesis test of a single proportion to determine if the proportion of late arrivals has increased:
8. What is the correct statement of null and alternative hypothesis?
A) H0: p < .18 and HA: p ³ .18
B) H0: p £ .18 and HA: p > .18
C) H0: p = .18 and HA: p ≠ .18
D) H0: p ³ .18 and HA: p < .18
E) H0: p £ .20 and HA: p > .20
9. What is the value of the calculated test statistic?
A) Z = 6.416 B) Z = 3.208 C) Z = -3.208
D) Z = -6.416 E) Z = 1.833
10. A research and development specialist for a company that produces a certain soft drink wants to determine if a 12-ounce can of a certain brand of soft drink contains 120 calories as the labeling indicates. Using a random sample of 9 cans, the specialist determined that the average calories per can is 124 with a standard deviation of 6 calories. At the .05 level of significance, is there sufficient evidence that the average calorie content of a 12-ounce can is different from 120 calories? Assume that the number of calories per can is normally distributed.
A) No, t = .667 and .667 is less than the critical value of 2.306.
B) No, t = 2 and 2 is less than the critical value of 2.306.
C) Yes, t = 2.5 and 2.5 is greater than the critical value of 2.262.
D) Yes, t = 2 and 2 is greater than the critical value of 1.833.
E) No, t = .667 and .667 is less than the critical value of 2.262.
Task D: A statistical test using ANOVA gives the following results:
SSTR = 6,750; SSE = 8000; nT = 20
H0: m 1 = m 2 = m 3 = m 4
Ha: Not all population means are equal
ANSWER QUESTIONS 11 THROUGH 15 BELOW.
11. The treatment mean square value (MSTR) is:
a. 400 b. 500 c. 1687.5 d. 2250
12. The error mean square (MSE) is:
a. 400 b. 500 c. 1687.5 d. 2250
13. The computed (i.e., observed) F-value is:
a. 0.22 b. 0.84 c. 4.22 d. 4.5
14. The null hypothesis is to be tested at the 5% level of significance. The critical F-value is:
a. 2.87 b. 3.24 c. 4.08 d. 8.7
15. The null hypothesis should:
a. not be rejected
b. be rejected
c. the test is inconclusive
d. none of the above answers are correct
Task E:
A company wants to measure the relationship between its employee productivity (measured in output/employee) and the number of employees. Sample data for the last four months are shown below. Use simple linear regression to estimate this relationship.
Independent Variable Dependent Variable
Number of Employees Employee Productivity
15 5
12 7
10 9
7 11
ANSWER QUESTIONS 16 THROUGH 19 BELOW.
16. The least squares estimate of the slope b1 is:
a. -0.7647 b. -0.13 c. 21.4 d. 16.41
17. The least squares estimate of the intercept b0 is:
a. -7.647 b. -0.13 c. 21.4 d. 16.41
18. The estimated employee productivity when the number of employees is 5 is:
a. 78 b. 12.59 c. 5.8 d. 32.6
19. If the sample covariance is -8.67; estimate the coefficient of correlation between the number of employees and employee productivity:
a. -0.997 b. 0.997 c. 1.23 d. 1.02
Task F:
Consumer Research is an independent agency that is collecting data on annual income (INCOME) and household size (SIZE), to predict annual credit card charges. It runs a regression analysis on the data and an incomplete MS Excel output is shown below.
ANSWER QUESTIONS 20 THROUGH 30 BELOW.
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Regression Statistics
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Multiple R
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0.88038239
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R Square
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Adjusted R Square
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Standard Error
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510.495493
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Observations
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ANOVA
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df
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SS
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MS
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F
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Significance F
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Regression
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2
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17960368.3
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3.31446E-07
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Residual
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20
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260605.648
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Total
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22
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Coefficients
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Standard Error
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t Stat
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P-value
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Lower 95%
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Intercept
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352.694714
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4.15578994
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0.00048872
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730.0172039
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INCOME
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25.062956
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8.47147285
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2.95851223
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0.00776734
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7.391781505
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SIZE
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408.400776
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71.808401
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1.447E-05
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258.6111461
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20. The sample size is:
a. 23 b. 22 c. 20 d. 21
21. The coefficient of determination is:
a. 0.88 b. 0.775 c. 0.92 d. -0.38
22. The Sum of Squares for Error (i.e., Residual) is:
a. 17960368.3 b. 5212112.97 c. 23172481.3 d. 260605.648
23. The Sum of Squares for Total (SST) is:
a. 17960368.3 b. 5212112.97 c. 23172481.3 d. 260605.648
24. The Mean Square for Regression is
a. 17960368.3 b. 5212112.97 c. 260605.648 d. 8980184.17
25. The observed or computed F-value is:
a. 34.459 b. 0.029 c. 3.445 d. 0.29
26. The hypothesis to be tested is:
H0: B1 = B2 = 0
Ha: At least one of the B is not equal to 0.
The hypothesis is to be tested at the 5% level of significance. The null hypothesis is:
a. not rejected
b. rejected
c. the test is inconclusive
d. none of the above answers are correct
27. The hypothesis to be tested is:
H0: B1 = 0
Ha: B1 + 0
The hypothesis is to be tested at the 1% level of significance. The null hypothesis is:
a. not rejected
b. rejected
c. the test is inconclusive
d. none of the above answers are correct
28. The estimate of the intercept b0 is:
a. 10010.2 b. 2810.3 c. 1465.5 d. 2641.5
29. The observed or computed t-stat (i.e., t-value) for the independent variable SIZE is:
a. 2.96 b. 3.445 c. 4.16 d. 5.687
30. What is the estimated annual credit charges if INCOME = 20, and SIZE = 3?
a. 9700 b. 12600 c. 3189 d. 5300
TaskG:
Last 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 number of students in each classification.
Freshmen 83
Sophomores 68
Juniors 85
Seniors 64
We want to know if there has been a significant change in the proportions of student classifications between the two years.
ANSWER QUESTIONS 31 THROUGH 34 BELOW.
31. The expected number of freshmen in this year is:
a. 83 b. 90 c. 30 d. 10
32. The number of degrees of freedom is:
a. 4 b. 2 c. 3 d. 1
33. The hypothesis is to be tested at the 5% level of significance. The critical chi-square value from the table equals:
a. 1.645 b. 1.96 c. 2.75 d. 7.815
34. If the chi-square value that is calculated equals 1.6615, then the null hypothesis is:
a. not rejected
b. rejected
c. the test is inconclusive
d. none of the above answers are correct
Task H: Use the following Excel Output to answer questions 35-39:
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Source
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Sum of Squares
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d.f.
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Between Groups
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213.88125
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3
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Within Groups
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11.208333
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20
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Total
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225.0895
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23
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35. Consider the above one-way ANOVA table. What is the treatment mean square?
A) 71.297 B) 0.5604 C) 1.297 D) 213.881 E) 9.7
36. Consider the above one-way ANOVA table. What is the mean square error?
A) 71.297 B) 0.5604 C) 1.297 D) 213.8810 E) 9.7
37. Consider the above one-way ANOVA table. How many groups (treatment levels) are included in the study?
A) 3 B) 4 C) 6 D) 20 E) 24
38. Consider the above one-way ANOVA table. If there are equal number of observations in each group, then each group (treatment level) consists of ______ observations.
A) 3 B) 4 C) 6 D) 20 E) 24
39. What is the critical F-value at an alpha of 0.05?
A) 3.1 B) 3.86 C) 14.17 D) 4.94 E) 8.66
Task I:
Use the following to answer questions 40-42:
The following results were obtained from a simple regression analysis:
Y^ = 37.2895 - (1.2024) * X
r = - 0.6774
40. For each unit change in X (independent variable), the estimated change in Y (dependent variable) is equal to:
A) -1.2024 B) 0.6774 C) 37.2895 D) 0.2934
41. When X (independent variable) is equal to zero, the estimated value of Y (dependent variable) is equal to:
A) -1.2024 B) 0.6774 C) 37.2895 D) 0.2934
42. __________ is the proportion of the variation explained by the simple linear regression model:
A) 0.8230 B) 0.6774 C) 0.4589 D) 0.2934 E) 37.2895
43. Given the following information about a hypothesis test of the difference between two means based on independent random samples, which one of the following is the correct rejection region at a significance level of .05?
HA: μA > μB, X‾1 = 12, X‾2 = 9, s1 = 4, s2 = 2, n1 = 13, n2 = 10.
A) Reject H0 if Z > 1.96
B) Reject H0 if Z > 1.645
C) Reject H0 if t > 1.721
D) Reject H0 if t > 2.08
E) Reject H0 if t > 1.734
44. Given the following information about a hypothesis test of the difference between two means based on independent random samples, what is the calculated value of the test statistic? Assume that the samples are obtained from normally distributed populations having equal variances
HA: μA > μB, X‾1 = 12, X‾2 = 9, s1 = 5, s2 = 3, n1 = 13, n2 = 10.
A) -1.674 B) 1.5 C) 2.823 D) 1.78 E) 1.063
Task J: Use the following to answer questions 45-46:
Consider the following partial computer output for a multiple regression model.
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Predictor
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Coefficient (bi)
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Standard Dev (sb)
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Constant
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99.3883
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X1
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-0.007207
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0.0031
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X2
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0.0011336
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0.00122
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X3
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0.9324
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0.373
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Analysis of Variance
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Source
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df
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SS
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Regression
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4
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32
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Error (residual)
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17
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8
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45. How many observations were taken?
A) 21 B) 17 C) 22 D) 20 E) 4
46. What is the value of R2?
A) 35% B) 76.95% C) 80% D) 77% E) 25%
Task K: Business travelers were asked to rate Miami Airport (on a scale of 1-10). Similarly business travelers were asked to rate Los Angeles airport. A hypothesis test (at alpha = 0.05) is conducted for any difference in the population means in the ratings. The Excel output is shown below. Use the following to answer questions 47- 48:
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t-Test: Two-Sample Assuming Unequal Variances
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Miami
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Los Angeles
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Mean
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6.34
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6.72
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Variance
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4.677959184
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5.63428571
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Observations
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50
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50
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Hypothesized Mean Difference
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0
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df
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97
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t Stat
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-0.836742811
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P(T<=t) one-tail
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0.202396923
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t Critical one-tail
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1.660714588
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P(T<=t) two-tail
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0.404793846
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t Critical two-tail
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1.984722076
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47. The Excel Output indicates that the null hypothesis is:
a. not rejected
b. rejected
c. the test is inconclusive
d. none of the above answers are correct
48. A 95% confidence interval of the difference between the mean ratings is:
a. - 0.52 to 1.25
b. 1.67 to 2.43
c. -0.51 to 1.27
d. -1.28 to 0.52
e. -2.43 to 1.67
49. A simple random sample of 100 observations was taken from a large population. The sample mean was 320 and the population standard deviation is 200. The standard error of the mean is:
a. 14.14 b. 22 c. 11 d. 20
50. Random samples of size 100 are taken from a population whose mean is 300 and standard deviation is 40. The expected value of x-bar and standard error of the mean are:
a. 300 and 40 b. 100 and 40 c. 300 and 4 d. 300 and 3
51. A population has a mean of 80 and a standard deviation of 10. A sample of 100 observations is taken. The probability that the sample mean is larger than 78 is:
a. 0.9772 b. 0.4772 c. 0.0228 d. 0.3413
52. A population has a mean of 180 and a variance of 1600. A sample of 100 observations is taken. The probability that the sample mean will be between 183 and 186 is:
a. 0.75 b. 0.1598 c. 0.4332 d. 0.2734
53. A population has a variance of 22. What should be the sample size so that there is a 95.4% probability that a sample mean will be within 3 units of the population mean:
a. 0.4772 b. 506 c. 10 d. 100
54. Random samples of size 40 are taken from an infinite population whose population proportion is 0.6. The standard error of the proportion is:
a. 0. 24 b. 0.006 c. 0.0774 d. 0.02445
55. A sample of 100 observations will be taken from an infinite population. The population proportion equals 0.2. The probability that the sample proportion will be more than 0.23 is:
a. 0.7734 b. 0.75 c. 0.2734 d. 0.2266
56. The t-value at a 95% confidence level and 25 degrees of freedom is:
a. 1.711 b. 2.060 c. 1.708 d. 2.064
57. A random sample of 16 statistics exams was taken. The average score in the sample was 76.2 with a variance of 144. The 99% confidence interval for the population average examination score is:
a. 49.67 to 102.72
b. 73.99 to 78.41
c. 67.36 to 85.04
d. 77.44 to 94.96
58. The sample size needed to provide a margin of error of 2 or less with a 95% probability when the population standard deviation equals 11 is:
a. 10 b. 11 c. 125 d. 117
Task I :
A local university wants to estimate the average time/week spent on computer laboratory terminals by each student. Data was collected for a sample of 81 students. The sample mean is 9 hours and standard deviation is 1.2 hours.
ANSWER QUESTIONS 49 THROUGH 51 BELOW.
59. The standard error of the mean is:
a. 7.5 b. 0.014 c. 0.16 d. 0.133
60. With 0.95 probability the margin of error is approximately:
a. 0.26 b. 1.96 c. 1.21 d. 1.64
61. The 95% confidence interval is:
a. 7.04 to 11.096 hours
b. 7.36 to 10.64 hours
c. 7.8 to 10.2 hours
d. 8.74 to 9.26 hours
62. The internal auditing staff of a local manufacturing company performs a sample audit each quarter to estimate the proportion of accounts that are delinquent (more than 90 days overdue). For this quarter, the auditing staff randomly selected 400 customer accounts and found that 80 of these accounts were delinquent. What is the 95% confidence interval for the proportion of all delinquent customer accounts at this manufacturing company?
A) .1608 to .2392
B) .1992 to .2008
C) .1671 to .2329
D) .1485 to .2515
E) .1714 to .2286
63. In a manufacturing process, we are interested in measuring the average length of a certain type of bolt. Past data indicates that the standard deviation is .25 inches. How many bolts should be sampled in order to make us 95% confident that the sample mean bolt length is within .02 inches of the true mean bolt length?
A) 25 B) 49 C) 423 D) 601 E) 1225
64. A plant quality control manager wishes to estimate the strength of a cable wire. A random sample of 64 wires is tested. The strength of the cable wire is measured in pounds per square inch. Based on the sample results, the sample mean is 238.4 pounds and the sample standard deviation is 40 pounds. What is the 99% confidence interval?
A) 227.96 to 248.84
B) 237.03 to 239.77
C) 225.53 to 251.28
D) 236.4 to 240.4
E) 229.21 to 247.59
65. According to a recent major survey, 70% of the respondents indicated that there is too much violence on TV. For a random sample of 40 people, what is the 95% confidence interval of the proportion of people who feel that there is too much violence on TV?
A) .70 ± .12
B) .70 ± .03
C) .70 ± .90
D) .70 ± .14
E) .70 ± .07
Task J: Based on the above portion of an Excel output, answer questions 56-58.
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Mean
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6.525
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Standard Error
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Median
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6.45
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Mode
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6
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Standard Deviation
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0.5437
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Sample Variance
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Count
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20
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Based on the above portion of an Excel output, answer questions 28-30.
66. The standard error is:
a. 0.5437 b. 0.2236 c. 0.0272 d. 0.1215
67. The Margin of Error at a 95% confidence level is:
a. 0.254 b. 1.138 c. 0.057 d. 0.468
68. The confidence interval is:
A) 6.45 ± 0.5437
B) 6.525 ± 0.057
C) 6.525 ± 0.254
D) 6.0 ± 1.138
E) 6.525 ± 0.468