ISDS 361BTest 1
Dr. Chen SAMPLE
INSTRUCTIONS:
- There are 23 multiple questions on the test.
- Data file is in the LANSCHOOL folder of you Drive C – DATA 1
- Fill in your section number, last name, first name and station number on the STUDENT worksheet
- Use Data file as your calculation paper
- Mark you answer on scantron
- Turn in test questions and scantron at the end of test
- University policy for academic dishonesty is to be strictly followed.
Lamps Life Time Problem
A certain brand of flood lamps has a lifetime that is normally distributed with a mean of 3,750 hours and a standard deviation of 300 hours.
- {Lamps Life Time} What proportion of these lamps will last for more than 4,000 hours?
- 0.2023
- 0.7977
- 0.9032
- -0.7977
- None of the above
- {Lamps Life Time} What lifetime should the manufacturer advertise for these lamps in order that only 2% of the lamps will burn out before the advertised lifetime? (round the number to nearest integer)
- 3134
- 3051
- 7401
- 3239
- None of the above
- The t criticalvalue for = 5% (one tail) and DF = 10 is
- 1.812
- 1.372
- 2.228
- 1.833
- None of the above
- Two independent samples of sizes 25 and 35 are randomly selected from two normal populations with equal variances. In order to test the difference between the population means, the test statistic is:
- a standard normal random variable
- t-distributed with 24 degrees of freedom
- t-distributed with 58 degrees of freedom
- t-distributed with 34 degrees of freedom
- None of the above
- In testing whether the means of two normal populations are equal, summary statistics computed for two independent samples are as follows:, , , , , and . Assume that the population variances are equal. Then, the standard error of the sampling distribution of the samplemean difference is equal to:
- 0.3255
- 1.2713
- 0.3189
- 1.1275
- None of the above
- In testing for differences between the means of two independent populations the null hypothesis is:
- None of the above
(Both a and b are correct)
- An attitude test which measures motivation for upward mobility was administered to a group of Japanese managers and another group of American managers. The test scores are summarized below.
American / Japanese
Sample Size / 200 / 100
Mean Test Score / 65.75 / 62.85
PopulationStandard Deviation / 11.00 / 9.00
Judging from the data collected which test would like to perform to determine if the average test scores of Japanese managers differs from the average test scores of American managers?
- z test for the difference of two population means
- t test for the difference of two population means assuming equal variances
- t test for the difference of two population means assuming unequal variances
- t test for difference of two paired data means
- None of the above
Coffee Breaks Problem
Do government employees take longer coffee breaks than private sector workers? That is a question that interested a management consultant. To examine the issue, he took a random sample of ten government employees and another random sample of ten private sector workers and measured the amount of time (in minutes) they spent in coffee breaks during the day. The results are listed below.
Government Employees / Private Sector Workers23 / 25
18 / 19
34 / 18
31 / 22
28 / 28
33 / 25
12 / 21
38 / 21
32 / 20
21 / 16
- {Coffee Breaks}Before testing difference in average time spent in coffee breaks, I perform an F-test for two sample variances as follows:
I conclude that the variances are not equal at 5% significance level because
- Government employees variance > Private sector worker’s variance
- F > F Critical one-tail (b is incorrect, it should be F Critical two-tail)
- One-tail P-value is smaller than 5%
- Twice of one-tail P-value is smaller that 5% (5pts)
- None of the above
- {Coffee Breaks}Is there enough evidence at the 5% significance level to support the consultant’s claim that government employees take longer coffee breaks than private sector workers? (Use worksheet COFFEE BREAK to perform test)
- Yes, the one tail P-value is 0.0382
- Yes, the one tail P-value is 0.0348
- No, the two tail P-value is 0.0764
- No, the two tail P-value is 0.0697
- None of the above
- Annual incomes (in $1,000s) of 10 sets of 30-year old twins are recorded as follows:
The 95% confidence interval for the difference in income between the twins is (Use worksheet TWINS to construct confidence interval):
- -2.19 to 8.98
- -12 to 7
- -8.03 to 0.13
- 0 to 4
- None of the above
- The test statistic of the single-factor ANOVA equals
- sum of squares for treatments divided by sum of squares for error
- sum of squares for error divided by sum of squares for treatments
- mean square for treatments divided by mean square for error
- mean square for error divided by mean square for treatments
- None of the above
- In single-factor ANOVA, suppose that there are four treatments with , , , and . Then the rejection region for this test at the 5% level of significance is
- Ftest
- Ftest
- Ftest
- Ftest
- None of the above
- A recent college graduate is in the process of deciding which one of three graduate schools he should apply to. He decides to judge the quality of the schools on the basis of the GMAT scores of those who are accepted into the school. A random sample of six students in each school produced the following GMAT scores. Assume that the data are normally distributed with equal variances:
School 1 / School 2 / School 3
650 / 520 / 590
620 / 550 / 510
600 / 600 / 600
580 / 630 / 500
710 / 600 / 490
690 / 650 / 530
The one-factor ANONA test concludes that means among 3 schools are NOT all equal. Use Fisher’s LSD at 5% significant level to determine if only School 1 differs from School 2 (Use worksheet GMAT for calculation):
- LSD = 49.65
- LSD = 60.37
- LSD = 76.3
- LSD = 90.07
- None of the above
- In order to examine the differences in ages of teachers among five school districts, an educational statistician took random samples of six teachers’ ages in each district. The data are listed below:
Can we infer at 5% significant level that the average teacher’s ages differ among the five school districts? (Use worksheet Teacher to perform test)
- No, because the P-value of ANOVA test is negative
- Yes, because the Ftest > Fcritical
- Yes, because MSTR > MSE
- Yes, because SSTR > SSE
- None of the above
- When you complete the following ANOVA table by filling in the blanks (identified by asterisks): (Use worksheetANOVAtable for calculation)
- Ftest = 1.56
- Ftest = 6.24
- Ftest = 0.20
- Not enough information to calculate Ftest
- None of the above
- A randomized block design experiment produced the following data: (Data is presented in worksheet Randomized Block)
- Treatment means differs only Correction!
- Block means differs only
- Both treatment and block means differ
- Neither treatment nor block means differs
- None of the above
- In the two-factor there are 4 levels for factor A, 5 levels for factor B, and 3 observations for each combination of factor A and factor B levels. The degree of freedom for the “error term” is:
- 12
- 59
- 6
- 40
- None of the above
Big Boy Problem: The following time series representing monthly sales of dishwashers at Big Boys Appliances over the past twelve months:
- {Big Boy Problem} What is the forecast over the next 4 months (total) assuming Big Boys uses a four month simple moving average?
- 110.08
- 113
- 112
- 110.5
- None of the above
- {Big Boy Problem}Big Boys is interested in selecting the technique which minimizes mean squared error, should it use a three month or a four month simple moving average.
- 3-month moving average
- 4-month moving average
- It makes no difference which technique is used
- Not enough information to calculate mean squared error
- None of the above
- {May Tag} Weekly sales of Maytag Neptune washing machines over the past sixteen weeks at Pacific Sales Appliance have been as follows:
To verify if the time series exhibit a linear trend, we will use
- t-test for two sample means assuming variances
- Test of sample means for paired data
- Linear regression test
- Single factor ANOVA test
- None of the above
- {May Tag} If the above time series exhibit a trend, using Linear Regression method what is the forecast of week 18?
- 17.73
- 15.25
- 17.05
- 14
- None of the above
- RDN’s sales of cable modem in San Mateo, California, for the months of January through April were as follows: January – 50, February - 80, March - 70, and April - 60. If exponential smoothing is used with a smoothing constant, alpha, of .20, what the forecast for May would be?
- 58.8
- 59.04
- 60
- 65
- None of the above
- A stationary forecasting model is appropriate for a time series which exhibits primarily:
- Trend
- Cyclical influences
- Seasonal components
- Random variation
- All of the above
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