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1. Assume that the data has a normal distribution and the number of observations is greater than fifty. Find the critical z value used to test a null hypothesis. ALPHA = 0.05 for a left-tailed test. (Points : 4) -1.96 Âą1.96 Âą1.645 -1.645 2. Find the value of the test statistic z using z = p^-p/vpq/n The claim is that the proportion of accidental deaths of the elderly attributable to residential falls is more than 0.10, and the sample statistics include n = 800 deaths of the elderly with 15% of them attributable to residential falls. (Points : 4) 3.96 -3.96 4.71 -4.71 3. Use the given information to find the P-value. Also, use a 0.05 significance level and state the conclusion about the null hypothesis (reject the null hypothesis or fail to reject the null hypothesis). The test statistic in a right-tailed test is z = 1.43. (Points : 4) 0.1528; fail to reject the null hypothesis 0.1528; reject the null hypothesis

0.0764; fail to reject the null hypothesis 0.0764; reject the null hypothesis 4. Find the number of successes x suggested by the given statement. Among 720 people selected randomly from among the residents of one city, 13.89% were found to be living below the official poverty line. (Points : 4) 100 101 104 99 5. Assume that you plan to use a significance level of alpha = 0.05 to test the claim that p1 = p2. Use the given sample sizes and numbers of successes to find the z test statistic for the hypothesis test. A random sampling of sixty pitchers from the National League and fifty-two pitchers from the American League showed that 19 National and 11 American League pitchers had E.R.A’s below 3.5. (Points : 4) z = 1.629 z = 1.253 z = 191.183 z = 15.457 6. Assume that you plan to use a significance level of alpha = 0.05 to test the claim that p1 = p2. Use the given sample sizes and numbers of successes to find the P-value for the hypothesis test. n1 = 50; n2 = 50 x1 = 8; x2 = 7 (Points : 4) 0.3897 0.6103 0.7794 0.2206 7. Construct the indicated confidence interval for the difference between population proportions . Assume that the samples are independent and that they have been randomly selected. In a random sample of 300 women, 49% favored stricter gun control legislation. In a random sample of 200 men, 28% favored stricter gun control legislation. Construct a 98% confidence interval for the difference between the population proportions p1 – p2. (Points : 4) 0.126 < p1 – p2 < 0.294 0.110 < p1 – p2 < 0.310 0.122 < p1 – p2 < 0.298 0.099 < p1 – p2 < 0.321 8. Construct the indicated confidence interval for the difference between the two population means. Assume that the two samples are independent simple random samples selected from normally distributed populations. Also assume that the population standard deviations are equal (sigma1 = sigma2), so that the standard error of the difference between means is obtained by pooling the sample variances. A researcher was interested in comparing the resting pulse rates of people who exercise regularly and people who do not exercise regularly. Independent simple random samples were obtained of 16 people who do not exercise regularly and 12 people who do exercise regularly. The resting pulse rate (in beats per minute) of each person was recorded. The summary statistics are as follows. Do not exercise Exercise xbar1 = 73.4 beats/min xbar2 = 69.7 beats/min

s1 = 10.3 beats/min s2 = 8.6 beats/min n1 = 16 n2 = 12 Construct a 90% confidence interval for the difference between the mean pulse rate of people who do not exercise regularly and the mean pulse rate of people who exercise regularly. (Points : 4) -0.34 beats/min < mu1 – mu2 < 7.74 beats/min -1.13 beats/min < mu1 – mu2 < 8.53 beats/min -3.22 beats/min < mu1 – mu2 < 10.62 beats/min -2.57 beats/min < mu1 – mu2 < 9.97 beats/min 9. The two data sets are dependent. Find d-bar to the nearest tenth. X 11.2 11.3 11.6 12.9 10.6 Y 11.1 12.6 12.9 10.7 13.3 (Points : 4) -0.8 -0.6 -0.7 -0.4 10. Given the linear correlation coefficient r and the sample size n, determine the critical values of r and use your finding to state whether or not the given r represents a significant linear correlation. Use a significance level of 0.05. r = 0.843, n = 5 (Points : 4) Critical values: r = ±0.878, significant linear correlation Critical values: r = ±0.950, no significant linear correlation Critical values: r = 0.950, significant linear correlation Critical values: r = ±0.878, no significant linear correlation 11. Find the value of the linear correlation coefficient r. The paired data below consist of the test scores of 6 randomly selected students and the number of hours they studied for the test. Hours 5 10 4 6 10 9 Score 64 86 69 86 59 87 (Points : 4) 0.224 -0.678 0.678 -0.224 12. Suppose you will perform a test to determine whether there is sufficient evidence to support a claim of a linear correlation between two variables. Find the critical values of r given the number of pairs of data n and the significance level alpha. n = 17, alpha = 0.05 (Points : 4) r = 0.497 r = ± 0.606 r = ± 0.482 r = 0.482

13. Use the given data to find the best predicted value of the response variable. Nine pairs of data yield r = 0.867 and the regression equation y = 19.4 + 0.93x. Also, ybar = 64.7. What is the best predicted value of y for x = 50? (Points : 4) 64.7 65.9 79.6 57.8 14. Use the given data to find the equation of the regression line. Round the final values to three significant digits, if necessary. Two different tests are designed to measure employee productivity and dexterity. Several employees are randomly selected and tested with these results. Productivity 23 25 28 21 21 25 26 30 34 36 Dexterity 49 53 59 42 47 53 55 63 67 75 (Points : 4) y-hat = 2.36 + 2.03x y-hat = 75.3 â€“ 0.329x y-hat = 5.05 + 1.91x y-hat = 10.7 + 1.53x 15. Use the given information to find the coefficient of determination. A regression equation is obtained for a collection of paired data. It is found that the total variation is 130.3, the explained variation is 79.3, and the unexplained variation is 51. Find the coefficient of determination. (Points : 4) 0.391 1.643 0.609 0.643 16. Use the computer display to answer the question. A collection of paired data consists of the number of years that students have studied Spanish and their scores on a Spanish language proficiency test. A computer program was used to obtain the least squares linear regression line and the computer output is shown below. Along with the paired sample data, the program was also given an x value of 2 (years of study) to be used for predicting test score. The regression equation is Score = 31.55 + 10.90 Years. Predictor Coef StDev T P Constant 31.55 6.360 4.96 0.000 Years 10.90 1.744 6.25 0.000 S = 5.651 R-Sq = 83.0% R-Sq (Adj) = 82.7% Predicted values Fit StDev Fit 95.0% CI 95.0% PI 53.35 3.168 (42.72, 63.98) (31.61, 75.09) What percentage of the total variation in test scores is unexplained by the linear relationship between years of study and test scores?

(Points : 4) 17.0% 8.9% 83.0% 82.7% 17. Find the explained variation for the paired data. The paired data below consists of heights and weights of 6 randomly selected adults. The equation of the regression line is y = -181.342 + 144.46x. Find the explained variation. x Height (meters) 1.61 1.72 1.78 1.80 1.67 1.88 y Weight (kg) 54 62 70 84 61 92 (Points : 4) 100.06 1079.5 979.44 1149.2 18. Find the unexplained variation for the paired data. The paired data below consists of heights and weights of 6 randomly selected adults. The equation of the regression line is y = -181.342 + 144.46x. Find the unexplained variation. x Height (meters) 1.61 1.72 1.78 1.80 1.67 1.88 y Weight (kg) 54 62 70 84 61 92 (Points : 4) 1079.5 979.44 100.06 119.3 19. Find the total variation for the paired data. The paired data below consists of heights and weights of 6 randomly selected adults. The equation of the regression line is y = -181.342 + 144.46x. Find the total variation. x Height (meters) 1.61 1.72 1.78 1.80 1.67 1.88 y Weight (kg) 54 62 70 84 61 92 (Points : 4) 100.06 1,119.3 979.44 1,079.5 20. Given below are the analysis of variance results from a Minitab display. Assume that you want to use a 0.05 significance level in testing the null hypothesis that the different samples come from populations with the same mean. Identify the value of the test statistic. Source DF SS MS F p Factor 3 13.500 4.500 5.17 0.011 Error 16 13.925 0.870 Total 19 27.425

(Points : 4) 0.011 5.17 4.500 13.500 21. Given below are the analysis of variance results from a Minitab display. Assume that you want to use a 0.05 significance level in testing the null hypothesis that the different samples come from populations with the same mean. Identify the p-value. Source DF SS MS F p Factor 3 13.500 4.500 5.17 0.011 Error 16 13.925 0.870 Total 19 27.425 (Points : 4) 4.500 5.17 0.870 0.011 22. Given below are the analysis of variance results from a Minitab display. Assume that you want to use a 0.05 significance level in testing the null hypothesis that the different samples come from populations with the same mean. What can you conclude about the equality of the population means? Source DF SS MS F p Factor 3 13.500 4.500 5.17 0.011 Error 16 13.925 0.870 Total 19 27.425 (Points : 4) Accept the null hypothesis since the p-value is greater than the significance level. Accept the null hypothesis since the p-value is less than the significance level. Reject the null hypothesis since the p-value is greater than the significance level. Reject the null hypothesis since the p-value is less than the significance level. 23. Provide an appropriate response. Which of the following nonparametric tests reaches a conclusion equivalent to the Mann-Whitney U test? (Points : 4) Wilcoxon rank-sum test sign test Wilcoxon signed-ranks test Kruskal-Wallis test 24. Find the critical value. Assume that the test is two-tailed and that n denotes the number of pairs of data. n =80, alpha = 0.05 (Points : 4) Âą 0.219 Âą 0.221

-0.221 0.221 25. When performing a rank correlation test, one alternative to using the Critical Values of Spearman’s Rank Correlation Coefficient table to find critical values is to compue them using this approximation rs = ± v t2/t2+n-2 where t is the t-score from the t Distribution table corresponding to n-2 degrees of freedom. Use this approximation to find critical values of rs for the case where n=7 and alpha = 0.05. (Points : 4) ±0.669 ±0.448 ±0.569 ±0.755

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