Binomial Distribution. Suppose that X has a binomial distribution with n = 50 and p = 0.6. Use Minitab to simulate 40 values of X.
MTB > random 40 c1;
SUBC > binomial 50 0.6.
Note: To find P(X < k) for any k > 0, use ‘cdf’ command; this works by typing:
MTB > cdf;
SUBC > binomial 50 0.6.
(a) What proportion of your values are less than 30? (b) What is the exact probability that X will be less than 30? (c) Find P(X < 28) and P(23 < X < 30) .
In: Math
4. ( Just true or false for each, no need for explanation,thank you)
a.If factors being studied cannot be controlled, the data are said to be observational.True or False
b.After rejecting the null hypothesis of equal treatments, a researcher decided to compute a 95 percent confidence interval for the difference between the mean of treatment 1 and mean of treatment 2 based on Tukey's procedure. At α = .05, if the confidence interval includes the value of zero, then we can reject the hypothesis that the two population means are equal.True or False
c.The error sum of squares measures the between-treatment variability.True or False
d.In one-way ANOVA, a large value of F results when the within-treatment variability is large compared to the between-treatment variability.True or False
e.In one-way ANOVA, other factors being equal, the further apart the treatment means are from each other, the more likely we are to reject the null hypothesis associated with the ANOVA F test. True or False
In: Math
In: Math
| HouseID | Age(years) | Size(sqft) | Bedrooms | Price($1,000) |
| 1 | 7 | 1580 | 3 | 132.0 |
| 2 | 8 | 1744 | 3 | 123.9 |
| 3 | 5 | 1863 | 3 | 159.1 |
| 4 | 7 | 1824 | 4 | 126.0 |
| 5 | 3 | 1924 | 4 | 128.3 |
| 6 | 10 | 1707 | 4 | 145.4 |
| 7 | 5 | 1898 | 3 | 126.1 |
| 8 | 5 | 2063 | 4 | 128.4 |
| 9 | 7 | 1641 | 3 | 147.4 |
| 10 | 7 | 1524 | 3 | 121.5 |
| 11 | 4 | 2144 | 3 | 167.7 |
| 12 | 9 | 1518 | 2 | 109.2 |
| 13 | 8 | 1645 | 3 | 132.6 |
| 14 | 5 | 2741 | 4 | 212.3 |
| 15 | 6 | 1718 | 3 | 111.8 |
| 16 | 7 | 1713 | 3 | 119.6 |
| 17 | 8 | 2240 | 4 | 162.0 |
| 18 | 6 | 1809 | 3 | 145.1 |
| 19 | 3 | 1588 | 4 | 124.4 |
| 20 | 9 | 1941 | 3 | 129.8 |
| 21 | 8 | 1783 | 4 | 127.4 |
| 22 | 6 | 2186 | 4 | 141.8 |
| 23 | 7 | 2018 | 4 | 131.3 |
| 24 | 4 | 2375 | 4 | 164.2 |
| 25 | 7 | 1801 | 4 | 127.9 |
| 26 | 5 | 2266 | 3 | 158.8 |
| 27 | 2 | 1891 | 3 | 131.2 |
| 28 | 6 | 2514 | 4 | 173.3 |
| 29 | 7 | 1738 | 3 | 121.4 |
| 30 | 6 | 2091 | 3 | 170.0 |
| 31 | 7 | 2495 | 4 | 185.6 |
| 32 | 5 | 1551 | 3 | 111.4 |
| 33 | 10 | 1814 | 4 | 134.9 |
| 34 | 9 | 1822 | 3 | 117.1 |
| 35 | 6 | 2289 | 4 | 186.8 |
| 36 | 8 | 2080 | 3 | 141.4 |
| 37 | 9 | 1758 | 3 | 137.6 |
| 38 | 9 | 2299 | 3 | 181.5 |
| 39 | 8 | 2190 | 3 | 139.2 |
| 40 | 8 | 1671 | 3 | 137.6 |
| 41 | 7 | 1544 | 3 | 137.4 |
| 42 | 6 | 1331 | 2 | 93.7 |
| 43 | 6 | 1386 | 3 | 86.5 |
| 44 | 4 | 2219 | 4 | 153.0 |
| 45 | 5 | 1793 | 3 | 129.4 |
| 46 | 8 | 1812 | 3 | 145.5 |
| 47 | 11 | 1499 | 3 | 124.2 |
| 48 | 4 | 1766 | 3 | 134.6 |
| 49 | 3 | 2060 | 3 | 162.5 |
| 50 | 9 | 1529 | 3 | 109.1 |
| 51 | 6 | 1276 | 2 | 97.9 |
| 52 | 7 | 2086 | 3 | 146.8 |
| 53 | 6 | 1395 | 2 | 95.1 |
| 54 | 4 | 1409 | 2 | 116.0 |
| 55 | 8 | 2009 | 3 | 155.3 |
| 56 | 7 | 1894 | 4 | 144.1 |
| 57 | 8 | 1190 | 2 | 88.8 |
| 58 | 5 | 1811 | 3 | 152.6 |
| 59 | 2 | 1689 | 3 | 146.5 |
| 60 | 7 | 1690 | 3 | 136.4 |
| 61 | 6 | 1685 | 3 | 125.2 |
| 62 | 6 | 1588 | 3 | 127.6 |
| 63 | 6 | 1730 | 3 | 143.6 |
| 64 | 5 | 1951 | 3 | 158.8 |
| 65 | 3 | 2040 | 3 | 139.9 |
| 66 | 8 | 1749 | 3 | 141.2 |
| 67 | 6 | 1111 | 2 | 86.7 |
| 68 | 5 | 1775 | 4 | 116.2 |
| 69 | 9 | 1478 | 2 | 100.7 |
| 70 | 10 | 2125 | 3 | 171.8 |
| 71 | 4 | 1596 | 3 | 135.1 |
| 72 | 5 | 1610 | 3 | 139.5 |
| 73 | 4 | 1724 | 2 | 151.4 |
| 74 | 4 | 1713 | 3 | 122.5 |
| 75 | 9 | 1490 | 3 | 115.3 |
| 76 | 10 | 1881 | 3 | 152.4 |
| 77 | 2 | 1868 | 3 | 162.0 |
| 78 | 8 | 2257 | 4 | 163.6 |
| 79 | 9 | 1614 | 4 | 145.3 |
| 80 | 5 | 1775 | 3 | 143.7 |
| 81 | 5 | 1770 | 3 | 129.3 |
| 82 | 2 | 2213 | 3 | 175.1 |
| 83 | 3 | 1743 | 3 | 152.2 |
| 84 | 6 | 2068 | 3 | 162.8 |
| 85 | 9 | 1705 | 3 | 126.7 |
| 86 | 6 | 1788 | 4 | 138.5 |
| 87 | 6 | 2334 | 4 | 180.1 |
| 88 | 5 | 1579 | 2 | 103.0 |
| 89 | 5 | 1478 | 2 | 105.1 |
| 90 | 4 | 2042 | 3 | 167.2 |
| 91 | 8 | 2263 | 3 | 158.8 |
| 92 | 9 | 1659 | 3 | 122.5 |
| 93 | 4 | 1978 | 4 | 131.1 |
| 94 | 4 | 1598 | 3 | 125.2 |
| 95 | 2 | 2523 | 4 | 197.9 |
| 96 | 7 | 1846 | 3 | 143.5 |
| 97 | 8 | 2292 | 3 | 163.2 |
| 98 | 8 | 1872 | 3 | 153.8 |
| 99 | 7 | 1954 | 3 | 146.1 |
| 100 | 5 | 1679 | 3 | 118.4 |
1). Which predictor variables are statistically significant at the 10% significance level?
2). What is the slope and p-value of the bedrooms variable?
3). What percentage of the variability in price is explained by this model?
In: Math
Answer the within-subjects ANOVA questions using the data below.
Use α = 0.01.
| 1 | 2 | 3 | 4 |
|---|---|---|---|
| 53 49 47 42 51 39 44 |
44 39 36 34 39 30 29 |
46 42 39 37 36 34 26 |
44 35 30 27 33 28 31 |
Make an interpretation based on the results.
At least one week differs on the GRE verbal score.No week is different on GRE verbal score.
e) Conduct Tukey's Post Hoc Test for the following
comparisons:
1 vs. 4: difference = ;
significant: ---Select--- Yes No
3 vs. 4: difference = ;
significant: ---Select--- Yes No
f) Conduct Scheffe's Post Hoc Test for the
following comparisons:
1 vs. 4: test statistic = ;
significant: ---Select--- Yes No
2 vs. 3: test statistic = ;
significant: ---Select--- Yes No
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How to do with TI-83 Plus?
In class we examined the linear relationship between shoe size and height for females. When looking at the male data, the following statistics were computed. For shoe size, the sample mean was 11.4 and the sample standard deviation was 1.42. For height, the sample mean was 71.1 and the sample standard deviation was 3. The sample correlation between shoe size and height was found to be r = 0.65. Find the regression equation for predicting height from shoe size, and use it to compute predicted values of height.
Compute the intercept of the regression line. Round your answer to 3 decimal places.
In: Math
An educational psychologist has developed a mediation technique
to reduce anxiety. The psychologist selected a sample of high
anxiety students that are asked to do the mediation at two therapy
sessions a week apart. The participants' anxiety is measured the
week before the first session and at each subsequent session. Below
are the anxiety scores for the participants. What can the
psychologist conclude with α= 0.05?
| before | session 1 | session 2 |
|---|---|---|
| 9 6 8 5 8 6 9 6 8 |
7 7 6 7 1 9 7 7 7 |
6 5 5 4 5 6 5 5 4 |
a) What is the appropriate test statistic?
---Select--- na one-way ANOVA within-subjects ANOVA two-way
ANOVA
b) Compute the appropriate test statistic(s) to
make a decision about H0.
critical value = ; test statistic =
Decision: ---Select--- Reject H0 Fail to reject H0
c) Compute the corresponding effect size(s) and
indicate magnitude(s).
η2 = ; ---Select--- na trivial
effect small effect medium effect large effect
d) Make an interpretation based on the
results.
At least one of the sessions differ on anxiety.None of the sessions differ on anxiety.
In: Math
Please use ONLY one Excel file to complete this case study, and use one spreadsheet for each problem.
Identify and remove the five cases corresponding to December revenue.
| Date | Wal Mart Revenue | CPI | Personal Consumption | Retail Sales Index | December |
| 10/31/03 | 12.452 | 562.4 | 7941071 | 302065 | 0 |
| 11/28/03 | 14.764 | 552.7 | 7868495 | 301337 | 0 |
| 12/30/03 | 23.106 | 552.1 | 7885264 | 357704 | 1 |
| 1/30/04 | 12.131 | 554.9 | 7977730 | 281463 | 0 |
| 2/27/04 | 13.628 | 557.9 | 8005878 | 282445 | 0 |
| 3/31/04 | 16.722 | 561.5 | 8070480 | 319107 | 0 |
| 4/29/04 | 13.98 | 563.2 | 8086579 | 315278 | 0 |
| 5/28/04 | 14.388 | 566.4 | 8196516 | 328499 | 0 |
| 6/30/04 | 18.111 | 568.2 | 8161271 | 321151 | 0 |
| 7/27/04 | 13.764 | 567.5 | 8235349 | 328025 | 0 |
| 8/27/04 | 14.296 | 567.6 | 8246121 | 326280 | 0 |
| 9/30/04 | 17.169 | 568.7 | 8313670 | 313444 | 0 |
| 10/29/04 | 13.915 | 571.9 | 8371605 | 319639 | 0 |
| 11/29/04 | 15.739 | 572.2 | 8410820 | 324067 | 0 |
| 12/31/04 | 26.177 | 570.1 | 8462026 | 386918 | 1 |
| 1/21/05 | 13.17 | 571.2 | 8469443 | 293027 | 0 |
| 2/24/05 | 15.139 | 574.5 | 8520687 | 294892 | 0 |
| 3/30/05 | 18.683 | 579.6 | 8568959 | 338969 | 0 |
| 4/29/05 | 14.829 | 582.9 | 8654352 | 335626 | 0 |
| 5/25/05 | 15.697 | 582.4 | 8644646 | 345400 | 0 |
| 6/28/05 | 20.23 | 582.6 | 8724753 | 351068 | 0 |
| 7/28/05 | 15.26 | 585.2 | 8833907 | 351887 | 0 |
| 8/26/05 | 15.709 | 588.2 | 8825450 | 355897 | 0 |
| 9/30/05 | 18.618 | 595.4 | 8882536 | 333652 | 0 |
| 10/31/05 | 15.397 | 596.7 | 8911627 | 336662 | 0 |
| 11/28/05 | 17.384 | 593.2 | 8916377 | 344441 | 0 |
| 12/30/05 | 27.92 | 589.4 | 8955472 | 406510 | 1 |
| 1/27/06 | 14.555 | 593.9 | 9034368 | 322222 | 0 |
| 2/23/06 | 18.684 | 595.2 | 9079246 | 318184 | 0 |
| 3/31/06 | 16.639 | 598.6 | 9123848 | 366989 | 0 |
| 4/28/06 | 20.17 | 603.5 | 9175181 | 357334 | 0 |
| 5/25/06 | 16.901 | 606.5 | 9238576 | 380085 | 0 |
| 6/30/06 | 21.47 | 607.8 | 9270505 | 373279 | 0 |
| 7/28/06 | 16.542 | 609.6 | 9338876 | 368611 | 0 |
| 8/29/06 | 16.98 | 610.9 | 9352650 | 382600 | 0 |
| 9/28/06 | 20.091 | 607.9 | 9348494 | 352686 | 0 |
| 10/20/06 | 16.583 | 604.6 | 9376027 | 354740 | 0 |
| 11/24/06 | 18.761 | 603.6 | 9410758 | 363468 | 0 |
| 12/29/06 | 28.795 | 604.5 | 9478531 | 424946 | 1 |
| 1/26/07 | 20.473 | 606.3 | 9540335 | 332797 | 0 |
| 2/23/07 | 21.922 | 619.4 | 9479239 | 335014 | 0 |
| 3/30/07 | 18.939 | 614.9 | 9583848 | 376491 | 0 |
| 4/27/07 | 22.47 | 619.8 | 9635181 | 366936 | 0 |
| 5/25/07 | 19.201 | 622.8 | 9698576 | 389687 | 0 |
| 6/29/07 | 23.77 | 623.9 | 9731285 | 382781 | 0 |
| 7/27/07 | 18.942 | 625.6 | 9799656 | 378113 | 0 |
| 8/31/07 | 19.38 | 626.9 | 9813630 | 392125 | 0 |
| 9/28/07 | 22.491 | 623.9 | 9809274 | 362211 | 0 |
| 10/26/07 | 18.983 | 621.6 | 9836807 | 364265 | 0 |
| 11/30/07 | 21.161 | 620.6 | 9870758 | 372970 | 0 |
| 12/28/07 | 31.245 | 622.5 | 9966331 | 434488 | 1 |
In: Math
Show that under the normality assumption, the F test is equivalent to the likelihood ratio test.
In: Math
A soft drink filling machine, when in perfect alignment, fills the bottles with 12 ounces of soft drink. A random sample of 36 bottles is selected, and the contents are measured. The sample yielded a mean content of 11.75 ounces with a standard deviation of 0.75 ounces.
a) Formulate the hypothesis to test to determine if the machine is in perfect adjustment.
b) Compute the value of the test statistic
c) Compute the p-value and give your conclusion regarding the adjustmentof the machine. Let a= 0.05.
On this problem the instructor has stated "For each question listed, explain how to get the correct answer. Think of this like an essay question. Or like you’re tutoring somebody. That’s what I’m really shooting for—for you to understand the material well enough to explain it to somebody else. If you can show me you can do that, you will get full credit.
So the answer has to be in essay form or comprehensive form. explaining the variables and how i got to the answer.
In: Math
Chapter 15 Discussion Group Question
Gender differences in dream content are well documented (see Winget & Kramer, 1979). Suppose a researcher studies aggression content in the dreams of men and women. Each participant reports his or her most recent dreams. Then each dream is judged by a panel of experts to have low, medium, or high aggression content. The observed frequencies are shown in the following matrix:
Aggression Content
| Low | medium | high | |
| female | 18 | 4 | 2 |
| male | 4 | 17 | 15 |
1. Write null and alternative hypotheses (in words and notation) for both ways of framing/interpreting this data (both 'Versions') for a Chi-Square Test of Independence.
2. Using Table 15.6 from your textbook as a model, create a frequency distribution matrix for this data set. Determine the observed and expected frequencies for this data set.
3. Write out the formula for degrees of freedom. Calculate degrees of freedom for this data set.
4. Determine the critical value from Appendix B in your textbook.
5. Compute a Chi-Square Test of Independence, use an α =.05. Include the Chi-Square result and the significance value (e.g., χ2 (1, n=200) = …., p < .05)
6. Would we use Cramer’s V or Phi-Coefficient to determine effect size for this data set? Use the test you just determined to find effect size for this data set.
7. Write the results and conclusions in an APA formatted results paragraph.Page 11 of the Help Guide should be helpful.
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The average age of a vehicle registered in Canada is about 97 months. If a random sample of 31 vehicles is selected, find the probability that the mean of their age is between 101 and 105 months. Assume the standard deviation for the population is 21.
In: Math
Let X1, X2, and X3 represent the times necessary to perform three successive repair tasks at a service facility. Suppose they are normal random variables with means of 50 minutes, 60 minutes, and 40 minutes, respectively. The standard deviations are 15 minutes, 20 minutes, and 10 minutes, respectively.
a) Suppose X1, X2, and X3 are independent. All three repairs must be completed on a given object. What is the mean and variance of the total repair time for this object?
b) Suppose X1, X2, and X3 are independent. All three repairs must be completed on a given object. Find the probability that the total repair time is less than 180 minutes.
c) Suppose that X1, X2, and X3 are dependent so that the covariance between X1 and X2 is -150, between X1 and X3 is 60, and between X2 and X3 is -45. If all three repairs must be completed on a given object, what is the mean and variance of the total repair time for this object?
In: Math
Overproduction of uric acid in the body can be an indication of cell breakdown. This may be an advance indication of illness such as gout, leukemia, or lymphoma.† Over a period of months, an adult male patient has taken eleven blood tests for uric acid. The mean concentration was x = 5.35 mg/dl. The distribution of uric acid in healthy adult males can be assumed to be normal, with σ = 1.89 mg/dl.
(a) Find a 95% confidence interval for the population mean concentration of uric acid in this patient's blood. What is the margin of error? (Round your answers to two decimal places.) lower limit upper limit margin of error
(b) What conditions are necessary for your calculations? (Select all that apply.) uniform distribution of uric acid n is large σ is unknown σ is known normal distribution of uric acid
(c) Interpret your results in the context of this problem. There is a 5% chance that the confidence interval is one of the intervals containing the population average uric acid level for this patient. The probability that this interval contains the true average uric acid level for this patient is 0.05. There is a 95% chance that the confidence interval is one of the intervals containing the population average uric acid level for this patient. The probability that this interval contains the true average uric acid level for this patient is 0.95. There is not enough information to make an interpretation.
(d) Find the sample size necessary for a 95% confidence level with maximal margin of error E = 1.14 for the mean concentration of uric acid in this patient's blood. (Round your answer up to the nearest whole number.) blood tests
In: Math