MIS131 Sections 7, 13
Sample Exam 1
Instructions: Download this word file and edit it to produce your exam results. Be sure to put your name on the disk. Upon completion, of the exam, save your work to your disk and turn it in. The data file for this sample exam is available from Dr. Freeman’s 131 web page.
Notes on cutting and pasting from Statgraphics to Word: Use the contol/insert keys together to copy a selected area of text to the Windows clipboard. Use shift/insert to paste into Word at the selected points. For graphs, first expand to fill the analysis window in statgraphics, then use Copy Graph from the file menu and copy the file to your desktop. Use insert | picture | file … to insert the graph into word. Grab a handle on the graph to reduce its size. Word will automatically place the figure.
- In order to estimate the mean of a population with a variance of 100 a sample of the population is to be taken.
NOTES: This analysis does not use statgraphics. You need to know the relationship between mean and variance, and the relationship between population standard deviation and the standard deviation of the estimate for the mean of the population.
- About how large must the sample be to give a 95% confidence interval of plus or minus 2 around the estimate of the mean?
Since the 95% confidence interval corresponds to +/- 2 sigma, the standard deviation of the estimate for the mean must be 1. The sample population has a variance of 100, therefore a standard deviation of 10. We need a large enough sample from the population to give us a factor of 1/10 on the population sigma, or 1/100 on the variance. Since the variance on the estimate of the mean is inversely proportional to the sample size, n = 100. OK, 101 is also acceptable.
- About how large must the sample be to give a 95% confidence interval of plus or minus 0.2 about the estimated mean? By the same reasoning, we need a factor of 1/100 on the standard deviation, or 1/10000 on the variance. Therefore, n = 10000. OK, 10001.
- Were the samples one and two in SD1.sf taken from the same population? Explain.
NOTES: You need Statgraphics. Use Compare | Two Samples | Two Sample Comparison. On the tabular options you have both a means comparison and a standard deviation comparison. Run one or both to determine your answer.
Answer: 95.0% confidence interval for the difference between the means
assuming equal variances: -0.383265 +/- 0.276379 [-0.659644,-0.106886]
t test to compare means
Null hypothesis: mean1 = mean2
Alt. hypothesis: mean1 NE mean2
assuming equal variances: t = -2.73467 P-value = 0.00681075
The StatAdvisor
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This option runs a t-test to compare the means of the two samples.
It also constructs confidence intervals or bounds for each mean and
for the difference between the means. Of particular interest is the
confidence interval for the difference between the means, which
extends from -0.659644 to -0.106886. Since the interval does not
contain the value 0.0, there is a statistically significant difference
between the means of the two samples at the 95.0% confidence level.
A t-test may also be used to test a specific hypothesis about the
difference between the means of the populations from which the two
samples come. In this case, the test has been constructed to
determine whether the difference between the two means equals 0.0
versus the alternative hypothesis that the difference does not equal
0.0. Since the computed P-value is less than 0.05, we can reject the
null hypothesis in favor of the alternative.
- Were the samples three and four in SD1.sf taken from the same population? Explain.
NOTES: See notes to 2.
Answer: three four
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Standard deviation 0.996125 2.33981
Variance 0.992265 5.47472
Df 99 99
Ratio of Variances = 0.181245
95.0% Confidence Intervals
Standard deviation of three: [0.874604,1.15717]
Standard deviation of four: [2.05437,2.7181]
Ratio of Variances: [0.121949,0.269372]
F-test to Compare Standard Deviations
Null hypothesis: sigma1 = sigma2
Alt. hypothesis: sigma1 NE sigma2
F = 0.181245 P-value = 0.0
The StatAdvisor
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This option runs an F-test to compare the variances of the two
samples. It also constructs confidence intervals or bounds for each
standard deviation and for the ratio of the variances. Of particular
interest is the confidence interval for the ratio of the variances,
which extends from 0.121949 to 0.269372. Since the interval does not
contain the value 1.0, there is a statistically significant difference
between the standard deviations of the two samples at the 95.0%
confidence level.
An F-test may also be used to test a specific hypothesis about the
standard deviations of the populations from which the two samples
come. In this case, the test has been constructed to determine
whether the ratio of the standard deviations equals 1.0 versus the
alternative hypothesis that the ratio does not equal 1.0. Since the
computed P-value is less than 0.05, we can reject the null hypothesis
in favor of the alternative.
- Is the continuous variable data in sample five in SD1.sf in statistical control?
NOTES: You need Statgraphics. Run Special | Quality Control | Variable Control Charts | xBar and R. Give enough information to support your answer.
Answer: The process is out of control. Point 11 exceeds the control upper limit on both the xbar and range charts.
- Is the proportional data in sample six in SD1.sf in statistical control with subgroup size of 500?
NOTES: You need Statgraphics. Run Special | Quality Control | Attrributse Control Charts | pchart. Give enough information to support your answer.
Answer: Yes. All the data is well within the control range.
- Which, if any, of samples seven, eight and/or nine in SD1.sf are related?
NOTES: Use Statgraphics. Run Describe | Categorical Data | Cross Tabulation and choose the Chi Squared test from the tabulation options.
Answer: Chi-Square Test
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Chi-Square Df P-Value
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25.34 16 0.0641
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Warning: some cell counts < 5.
The StatAdvisor
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The chi-square test performs a hypothesis test to determine whether
or not to reject the idea that the row and column classifications are
independent. Since the P-value is less than 0.10, we can reject the
hypothesis that rows and columns are independent at the 90% confidence
level. Therefore, the observed value of Seven for a particular
case is related to its value for Eight.
Chi-Square Test
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Chi-Square Df P-Value
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19.41 16 0.2479
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Warning: some cell counts < 5.
The StatAdvisor
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The chi-square test performs a hypothesis test to determine whether
or not to reject the idea that the row and column classifications are
independent. Since the P-value is greater than or equal to 0.10, we
cannot reject the hypothesis that rows and columns are independent.
Therefore, the observed value of Seven for a particular case may bear
no relation to its value for Nine.
Chi-Square Test
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Chi-Square Df P-Value
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16.48 16 0.4199
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Warning: some cell counts < 5.
The StatAdvisor
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The chi-square test performs a hypothesis test to determine whether
or not to reject the idea that the row and column classifications are
independent. Since the P-value is greater than or equal to 0.10, we
cannot reject the hypothesis that rows and columns are independent.
Therefore, the observed value of Eight for a particular case may bear
no relation to its value for Nine.
- The following questions refer to samples ten and eleven in SD1.sf
NOTES: Use Statgraphics. Run Relate | Simple Regression
- What portion of the variation in sample ten can be explained by the variation in sample eleven?
Answer: The R-Squared statistic indicates that the model as fitted explains
71.4812% of the variability in Ten.
- Is the dependence statistically significant?
Answer: Since the P-value in the ANOVA table is less than 0.01, there is a
statistically significant relationship between Ten and Eleven at the
99% confidence level.
- Describe the dependence.
Answer: The output shows the results of fitting a linear model to describe
the relationship between Ten and Eleven. The equation of the fitted
model is
Ten = 2105.02 + 9.7348*Eleven
- Sample 12 is the number of flight departures delayed per day at Megalopolis International Airport over a one month period.
NOTES: Use Statgraphics. Run Special | Quality Control | Attributes Control Charts | c Chart.. . Provide enough information it support your answer.
- Is the airport departure process in statistical control? Assume 600 flights per day.
Answer:
The departure process is out of control by inspection. Points 13 and 26 are beyond the control upper limit.
- A hurricane struck Megalopolis on day 13. Does removal of this specific cause bring the departure management under statistical control?
Answer:
Removal of specific cause due to 13 does not restore the process to being in control. Point 26 is still above the control range.