1. (Use BMD data) Researchers studied the bone mineral density (BMD) and depression of menaged 65 to 92 years. At the 1% significance level, do the data in the table (taken from a randomsample) provide sufficient evidence to conclude that BMD and depression are statisticallydependent for elderly men? The historical study shows that using Chi-square test is appropriateAВCD1DepressedNot_depressed2 Osteoporitic1373Low_BMD71530here.Normal951292а.State the null and alternate hypotheses.789.b.Find the value of the chi-square statistic for the sample. (Round the test statistic to three decimalplaces.)1011121314C.Find the P-value of the sample test statistic. (Round to four decimal places.)15161718d. Interpret your conclusion in the context of the application.1920Insert your R output here:+е.

(Since it has been not mentioned which question so, as per our protocol we shall only answer the…

1. (Use BMD data) Researchers studied the bone mineral density (BMD) and depression of men aged 65 to 92 years. At the 1% significance level, do the data in the table (taken from a random sample) provide sufficient evidence to conclude that BMD and depression are statistically dependent for elderly men? The historical study shows that using Chi-square test is appropriate A C D Depressed Not_depressed 2 Osteoporitic Low_BMD 37 3. 71 530 here. 4 Normal 95 1292 а. State the null and alternate hypotheses. 9 b. Find the value of the chi-square statistic for the sample. (Round the test statistic to three decimal places.) 10 11 12 13 14 c. Find the P-value of the sample test statistic. (Round to four decimal places.) 15 16

1. (Use BMD data) Researchers studied the bone mineral density (BMD) and depression of men aged 65 to 92 years. At the 1% significance level, do the data in the table (taken from a random sample) provide sufficient evidence to conclude that BMD and depression are statistically dependent for elderly men? The historical study shows that using Chi-square test is appropriate A C D Depressed Not_depressed 2 Osteoporitic Low_BMD 37 3. 71 530 here. 4 Normal 95 1292 а. State the null and alternate hypotheses. 9 b. Find the value of the chi-square statistic for the sample. (Round the test statistic to three decimal places.) 10 11 12 13 14 c. Find the P-value of the sample test statistic. (Round to four decimal places.) 15 16

MATLAB: An Introduction with Applications

6th Edition

ISBN:9781119256830

Author:Amos Gilat

Publisher:Amos Gilat

Chapter1: Starting With Matlab

Section: Chapter Questions

Problem 1P

See similar textbooks

Similar questions

also z test and p value
What is the test statistic and p-value? Because the P-value is ▼ greater less than the significance level 0.05, there ▼ is not is sufficient evidence to support the claim that there is a linear correlation between selling price (in hundred thousands) and the list price (in hundred thousands) of homes sold for a significance level of α=0.05.
The leader of two postpartum women’s support groups is interested in the depression levels of the women in her groups. She administers the Center for Epidemiologic Studies Depression Scale (CES-D) screening test to the members of her groups. The CES-D is a 20-question self-test that measures depressive feelings and behaviors during the previous week. The mean depression level from the screening test for the 10 women in the first group is μ₁ = 16; the mean depression level for the 14 women in the second group is μ₂ = 10. Without calculating the weighted mean for the combined group, you know that the weighted mean is:
can you explain how these scores were found, based on the statistics provided they dont seem to match up.' There is a significant interaction effect between employment status and substance use on stress levels. Two-Way ANOVA result: F(3,470) = 5.67, p = 0.001' I
Weather-Related Deaths Review Exercise 5 involved weather-related U.S. deaths. Among the 450 deaths included in that exercise, 320 are males. Use a 0.05 significance level to test the claim that among those who die in weather-related deaths, the percentage of males is equal to 50%. Provide an explanation for the results.
it y" 10. A dbly līne will be shut down for maintenance if the defect rate exceeds 2.3%. Suppose you adopt a 5% significance level and' take compute the proportion that are defective. For what values of the sample proportion will the assembly line be shut down? a random sample 200 items off the assembly line and 0.06 0.05 0.04 0.02 Back Net
A survey was conducted to investigate the relationship between gender (male and females) and sector of employment (private, government and academia). Using the information provided, does a relationship exist between gender and employment sector at the 5% significance level? If the Chi-square test statistics = 0.529, what conclusion can be made? A. Since Chi-square test statistics > Chi-square critical value, do not reject H0 B. Since Chi-square test statistics < Chi-square critical value, do not reject H0 C. Since Chi-square test statistics < Chi-square critical value, Reject H0 D. Since Chi-square test statistics > Chi-square critical value, Reject H0
int onodqolot o gnnu noiesup examine profit as a percentage of stockholder equity. A random sample of 31 retail stocks such as How profitable are different sectors of the stock market? One way to answer such a question is to Toys 'R' Us, Best Buy, and Gap was studied for x1, profit as a percentage of stockholder equity. The result was sample mean of 14.7. A random sample of 36 utility (gas and electric) stocks such 133 80 113 .s Boston Edison, Wisconsin Energy, and Texas Utilities was studied for x2, profit as a nercentage of stockholder equity. The result was a sample mean of 9.8. Assume that gi = 3.4 and 02 = 3.8. N, = 31 >X,=TH:1,0,=3.4, n2= 36, E,=9.8,0a C.L:0.95 At the 95% level of confidence, does it appear that the profit as a percentage of stockholder equity for retail stocks is higher than that for utility stocks? (Show your confidence in interval notation and show your interpretation) 2 Sampz %3D 3.1751 < M,- Azs 6.6243 We are 95% level Conzidence that the Profit Jol letail…
Please help! I need this to graduate, thank you so much.
This research was conducted with Korean adolescents in the Los Angeles area. Would you be willing to generalize the results of this study to Korean adolescents who live in other regions of the country? Why or why not.
Taxi Brand A Brand B1 38,300 39,8002 37,700 38,1003 42,200 43,0004 42,300 41,6005 48,200 47,0006 36,100 40,4007 35,300 36,4008 44,600 45,900
Is the proportion of wildfires caused by humans in the south different from the proportion of wildfires caused by humans in the west? 394 of the 533 randomly selected wildfires looked at in the south were caused by humans while 388 of the 554 randomly selected wildfires looked at the west were caused by humans. What can be concluded at the = 0.01 level of significance? For this study, we should use Select an answer z-test for the difference between two population proportions t-test for the difference between two dependent population means z-test for a population proportion t-test for the difference between two independent population means t-test for a population mean The null and alternative hypotheses would be: Select an answer p1 μ1 Select an answer > ≠ < = Select an answer μ2 p2 (please enter a decimal) Select an answer p1 μ1 Select an answer > = < ≠ Select an answer p2 μ2 (Please enter a decimal) The test statistic ? z t = (please show your…