Gender - Washington State University



STAT412Exam 2Practice QuestionsQuestions 1 to 3: In the General Social Survey, respondents were asked what they thought was most important to get ahead: hard work, lucky breaks, or both. Minitab output for 1026 respondents, by gender, is shown below:Expected counts are printed below observed counts Male Female Total Hard work 284 393 677 292.31 384.69 Lucky breaks 84 121 205 88.51 116.49 Both 75 69 144 62.18 81.82 Total 443 583 1026Chi-Sq = 0.236 + 0.180 + 0.230 + 0.175 + 2.645 + 2.010 = 5.476P-Value = 0.0651. State the research hypothesis.A) There is a relationship between gender and opinion on what is important to get ahead in the sample.B) There is no relationship between gender and opinion on what is important to get ahead in the sample.C) There is a relationship between gender and opinion on what is important to get ahead in the population.D) There is no relationship between gender and opinion on what is important to get ahead in the population.2. What are the degrees of freedom for the test statistic?A) 1B) 2C) 3D) 43. At a significance level of .05, what is the conclusion?A) Reject the null hypothesis and conclude there is no relationship between the variables.B) Reject the null hypothesis and conclude there is a relationship between the variables.C) Do not reject the null hypothesis and conclude the evidence is not strong enough to show a relationship between the two variables.D) Do not reject the null hypothesis and conclude there is a relationship between the variables.4. Ninety people with high cholesterol are randomly divided into three groups of thirty, and a different treatment program for decreasing cholesterol is assigned to each group. The response variable is the change in cholesterol level after two months of treatment. An analysis of variance F-test will be used to compare the three treatments. What null hypothesis is tested by this F-test?A) The sample variances are equal for the three treatment groups. B) The population variances are equal for the three treatments.C) The sample means are equal for the three treatment groups.D) The population means are equal for the three treatments5. A null hypothesis is that the probability is 0.7 that a new drug will provide relief in a randomly selected patient. The alternative is that the probability of relief is greater than 0.7. Suppose the treatment is used on 500 patients and there are 380 successes. How would a p-value be calculated in this situation? A) Find the chance of 380 or more successes, calculated assuming that ? is greater than 0.7.B) Find the chance of 380 or more successes, calculated assuming ? is equal to 0.7.C) Find the chance of fewer than 380 successes, calculated assuming that ? is greater than 0.7.D) Find the chance of fewer than 380 successes, calculated assuming ? is equal to 0.7.6. A Tukey’s multiple comparison is performed to compare 5 population means. How many confidence intervals will be obtained?A) 10B) 5C) 20D) 157. Table 1 below provides a relative frequency distribution for (reported) violent crimes in 1995. For instance, in 1995, 32.3% of violent crimes were robberies. A random sample of 500 violent-crime reports from last year yielded the frequency distribution shown in Table 2. The researcher was interested in determining if the data provide sufficient evidence to conclude that last year’s distribution of violent crimes has changed from the 1995 distribution. What test should be used? Table 1 Table 2 Distribution of violent crimes Sample results for 500 randomly selected in the United States, 1995 violent-crime reports last yearType of Violent CrimeRelative FrequencyType of Violent Crime FrequencyMurder0.012Murder9Forcible Rape0.054Forcible Rape26Robbery0.323Robbery144Agg. Assault0.611Agg. Assault3211.00500A) One-Factor ANOVAB) Two-Factor ANOVAC) Chi-Square test of IndependenceD) Chi-Square Goodness of Fit Test8. If we are performing a hypothesis test of H0: ?=.6 vs. Ha: ??>.6, the probability of rejecting H0 if ???? will be ________ the probability of rejecting H0 if ????.A) greater thanB) less thanC) equal toD) not comparable to9. For a survey of 900 college students, the following table gives mean classes missed per week classified by gender and whether or not students are in a Greek organization (sorority/fraternity).Greek organization member?GenderNoYesFemale1.231.24Male1.672.13Based on the means given in the table, it appears that gender and Greek organization membership (no or yes) areA) not interacting variables because members of Greek organization miss more classes per week than non-members do, regardless of gender.B) interacting variables because the difference between non-members and members of Greek organizations is greater for males than it is for females.C) interacting variables because males miss more classes than females.D) not interacting variables because males miss more classes than females regardless of Greek membership.10. A researcher is interested in estimating the proportion of voters who favor a tax on e-commerce. Based on a sample of 250 people, she obtains the following 99% confidence interval for the population proportion ?:0.113 < ? < 0.171Which of the statements below is a valid interpretation of this confidence interval?A) There is a 99% chance that ? lies between 0.113 and 0.171B) If many different samples of size 250 were selected and, based on each sample, a confidence interval were constructed, 99% of the time the value of ? would lie between 0.113 and 0.171C) The method used to get the interval from .113 to .171, when used over and over on different samples, produces intervals which include the population proportion 99% of the time in the long run. D) If 100 different samples of size 250 were selected and, based on each sample, a confidence interval were constructed, exactly 99 of these confidence intervals would contain the value of ?.11. Suppose the present success rate in the treatment of a particular psychiatric disorder is 65%. A research group hopes to demonstrate that the success rate of a new treatment will be better than this standard. Which of the following describes a type 1 error for this problem?A) Claiming that the success rate of the new treatment is greater than 65% when really it isn't.B) Failing to decide that the success rate is greater than 65% when actually it is.C) Using a one-sided alternative hypothesis when a two-sided alternative should have been used.D) Using a two-sided alternative hypothesis when a one-sided alternative should have been used.12. A student survey was done to study the relationship between gender and favorite television program watched on Sunday mornings (sports, news, or other). The chi-square test statistic was 8. What is the p-value or p-value range?A) .005 < p-value < 0.01B) .01 < p-value < .025C) .025 < p-value < .05D) None of the above13. The given observations are tomato yields (kg/plot) for four different levels of electrical conductivity (EC) of the soil. Chosen EC levels were 1.6, 3.8, 6.0, and 10.2 nmhos/cm.EC levelYield1.659.553.356.863.13.855.259.152.854.56.051.748.853.949.010.244.648.541.047.3 Level N Mean StDev 1.6 4 58.175 4.150 3.8 4 55.400 2.665 6.0 4 50.850 2.426 10.2 4 45.350 3.327 Calculate the value of the test statistic that would be used for testing H0: ?1.6=?3.8=?6.0=?10.2 vs. Ha: At least two of the population means are different.A) 6.35B) 12.20C) 24.15D) 40.0814. A survey claims that 9 out of 10 doctors recommend aspirin for their patients with headaches. To test this claim against the alternative that the actual proportion of doctors who recommend aspirin is less than 0.90, a random sample of 100 doctors was selected. Suppose that the test statistic is – 2.20. Can we conclude that H0 should be rejected at the (a) = 0.10, (b) = 0.05, and (c) = 0.01 level of significance?A) (a) yes; (b) yes; (c) yesB) (a) no; (b) no; (c) noC) (a) no; (b) no; (c) yesD) (a) yes; (b) yes; (c) no15. A sample of n=168 students was asked, “Do you believe in love at first sight?” The choices (below) show confidence intervals, in scrambled order, for 90%, 95%, 98%, and 99% confidence intervals for the population proportion who would answer yes. Which choice gives the 98% confidence interval? A) .56 to .68B) .52 to .72C) .53 to .71D) .55 to .6916. Tukey’s multiple comparison procedure is applied to compare the lifetimes of flashlight batteries of three different brands. Independent random samples of batteries of the three different brands yielded the following lifetimes in hours.In the table below are confidence intervals constructed using Tukey’s procedure. State which population means can be declared different. A) ?1 and ?2 B) ?1 and ?3 C) ?2 and ?3D) None17. In a two-factor ANOVA problem, there are 4 levels of factor A, 5 levels of factor B, and 2 observations (replications) for each combination of levels of the two factors. Then, the number of treatments in this experiment is A) 20 B) 11 C) 10 D) 40 18. Of 369 randomly selected medical students, 23 said that they planned to work in a rural community. Construct a 95% confidence interval for the percentage of all medical students who plan to work in a rural community.A) (4.66%, 7.80%)B) (2.99%, 9.47%)C) (3.30%, 9.17%)D) (3.77%, 8.70%)Problems:19. Use a significance level of 0.01 to test the claim that workplace accidents are distributed on workdays as follows: Monday 25%, Tuesday: 15%, Wednesday: 15%, Thursday: 15%, and Friday: 30%. In a study of 100 workplace accidents, 21 occurred on a Monday, 13 occurred on a Tuesday, 15 occurred on a Wednesday, 16 occurred on a Thursday, and 35 occurred on a Friday. H0: _____________Ha: _____________?=.01Specify rejection region: Calculate test statistic: State your Conclusion:Approximate p-value: 20. Consider the accompanying 2 2 table displaying the sample proportions that fell in the various combinations of categories (e.g., 18% of those in the sample were in the first category of both factors).121.18.402.07.35What is the smallest sample size n for which these observed proportions would result in rejection of the independence hypothesis? Use ?=.01.21. An agricultural scientist wants to determine how the type of fertilizer and the type of soil affect the yield of oranges in an orange grove. He has two types of fertilizer and three types of soil. For each of the combinations of fertilizer and soil, the scientist plants four stands of trees, and measures the yield of oranges (in tons per acre) from each stand. The data are shown in the following table.Soil Type 1Soil Type 2Soil Type 3Fertilizer A 20 24 22 922 22 27 2529 25 23 28Fertilizer B23 23 30 2423 29 31 3225 29 30 31Using the Minitab Output below, analyze the data using ?=.05. Also include a profile plot.Source DF SS MS F PFERTILIZER 1 121.50 121.500 7.84 0.012SOIL TYPE 2 141.75 70.875 4.57 0.025Interaction 2 14.25 7.125 0.46 0.639Error 18 279.00 15.500Total 23 556.50Tukey 95.0% Simultaneous Confidence IntervalsResponse Variable YIELD OF ORANGESAll Pairwise Comparisons among Levels of FERTILIZERFERTILIZER = A subtracted from:FERTILIZER Lower Center Upper ----+---------+---------+---------+--B 1.123 4.500 7.877 (----------------*---------------) ----+---------+---------+---------+-- 2.0 4.0 6.0 8.0Tukey 95.0% Simultaneous Confidence IntervalsResponse Variable YIELD OF ORANGESAll Pairwise Comparisons among Levels of SOIL TYPESOIL TYPE = 1 subtracted from:SOILTYPE Lower Center Upper +---------+---------+---------+------2 -0.5249 4.500 9.525 (-----------*------------)3 0.6001 5.625 10.650 (-----------*------------) +---------+---------+---------+------ -4.0 0.0 4.0 8.0SOIL TYPE = 2 subtracted from:SOILTYPE Lower Center Upper +---------+---------+---------+------3 -3.900 1.125 6.150 (------------*-----------) +---------+---------+---------+------ -4.0 0.0 4.0 8.0 ................
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