MATLAB: An Introduction with Applications
6th Edition
ISBN: 9781119256830
Author: Amos Gilat
Publisher: John Wiley & Sons Inc
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- 15. Research results suggest that IQ scores for boys Plomin, 2006). A typical study looking at 10-year-old are more variable than IQ scores for girls (Arden & size would be reported in a journal article. children classifies participants by gender and by low. average, or high IQ. Following are representing the research results. Do the data indicate a significant difference between the frequency distri- butions for males and females? Test at the .05 level of significance and describe the difference. hypothetical data 1 IQ Low Average High Вoys 18 42 20 80 Girls 12 54 14 80 n = 160 Int Avaarrow_forwardQ.7arrow_forwardDoes college major depend on gender? Researchers ask a group of college students about their majors. They also group the students by their gender. The data are recorded in the contingency table below, and a chi-square Test of Independence at the 5% significance level is performed. Arts Humanities Sciences Row Total Men 11 5 19 35 Women 11 22 9 42 Column Total 22 27 28 77 (a) The null and alternative hypotheses are: H0: The two variables are independent, so gender does not affect college major. Ha: The two variables are dependent, so gender does affect college major. (b) Compute the test statistic, rounded to one decimal place. (Use expected frequencies that are also rounded to one decimal place.)arrow_forward
- Winning team data were collected for teams in different sports, with the results given in the accompanying table. Use a 0.05 significance level to test the claim that home/visitor wins are independent of the sport. Given that among the four sports included here, baseball is the only sport in which the home team can modify field dimensions to favor its own players, does it appear that baseball teams are effective in using this advantage? Click the icon to view the data table. Determine the null and alternative hypotheses. Choose the correct answer below. OA. Ho: Basketball games are more likely to win at home than any other sport. H₁: Basketball games are not more likely to win at home than any other sport. OB. Ho: Basketball games are not more likely to win at home than any other sport. H₁: Basketball games are more likely to win at home than any other sport. OC. Ho: The home/visitor win is independent of the sport. H₁: The home/visitor win is not independent of the sport. O D. Ho: The…arrow_forward4 4G I. moodle.unizwa.edu.om a | Quiz3 Time left 0:00:21 Question 3 Answer saved Marked out of 2.00 P Flag question Which one of the following choices describes a problem for which an analysis of variance would be appropriate? O A. Comparing the proportion of successes for three different treatments of anxiety. Each treatment is tried on 100 patients O B. Analyzing the relationship between high school GPA and college GPA O . Analyzing the relationship between gender and opinion about capital punishment (favor or oppose). O D. Comparing the mean birth weights of newborn babies for three different hospitals. O E. None of these Clear my choice Previous page Next page 1 2w-ANOVA(3) Jump to... Quiz navigation 7 +arrow_forwardAssume that matched pairs of data from a poll of Internet users who were asked if they make travel plans through the Internet result in 373 positive signs, 333 negative signs, and 22 ties when the value of the second variable is subtracted from the corresponding value of the first variable. Use the sign test with a 0.10 significance level to test the null hypothesis of no difference. Let ₁ denote the median of the first variable and 12 denote the median of the second variable. Wha are the null and alternative hypotheses? A. Ho: 1₁ #1₂ H₁:1₁ =11₂ C. Ho: 1₁ =1₂ H₁:1₁ <1₂ Find the test statistic. Test statistic = needed.) OB. Ho: 1₁ 2112 H₁:1₁ <1₂ D. Ho: 1₁ =112 H₁:1₁ #1₂ (Round to two decimal places asarrow_forward
- The following table contains observed frequencies for a sample of 240. RowVariable Column Variable A B C P 20 30 20 Q 30 60 25 R 10 15 30 Test for independence of the row and column variables using ? = 0.05. 1) State the null and alternative hypotheses: A) H0: The column variable is independent of the row variable. Ha: The column variable is not independent of the row variable. B) H0: Variable P is not independent of variables Q and R. Ha: Variable P is independent of variables Q and R. C) H0: The column variable is not independent of the row variable. Ha: The column variable is independent of the row variable. D) H0: Variable P is independent of variables Q and R. Ha: Variable P is not independent of variables Q and R. 2)Find the value of the test statistic. (Round your answer to three decimal places.) Find the p-value. (Round your answer to four decimal places.) p-value = 3)State your conclusion: A) Do not reject H0. We cannot conclude that the column…arrow_forwardYou may need to use the appropriate technology to answer this question. Submitted The following table contains observed frequencies for a sample of 200. Column Variable Row Variable A B C P 20 44 50 Q 30 26 30 Test for independence of the row and column variables using a = 0.05. State the null and alternative hypotheses. Ho: The column variable is independent of the row variable. H: The column variable is not independent of the row variable. Ho: Variable P is not independent of variable Q. H: Variable P is independent of variable Q. Ho: Variable P is independent of variable Q. H: Variable P is not independent of variable Q. : The column variable is not independent of the row variable. Ho: : The column variable is independent of the row variable. Find the value of the test statistic. (Round your answer to three decimal places.) Find the p-value. (Round your answer to four decimal places.) p-value %3D State your conclusion. Reject Ho. We conclude that the column and row variables are…arrow_forwardTest the claim that the proportion of men who own cats is significantly different than the proportion of women who own cats at the 0.02 significance level. The null and alternative hypothesis would be: Но: HF Ho: UM = µF Ho:PM — pF Ho:рм PF Ho: им — MF Ho:; PF Нi: рм > иF Hi: рм # pF Hi:рм рF Hi: рм < pF Hi:рм # pғ The test is: right-tailed two-tailed left-tailed Based on a sample of 80 men, 25% owned cats Based on a sample of 40 women, 50% owned cats The test statistic is: (to 2 decimals) The p-value is: (to 2 decimals) Based on this we: O Reject the null hypothesis O Fail to reject the null hypothesisarrow_forward
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