Question

Need help with my statistics homework

QUESTION 2 In a survey performed at the University, it was found that 46 out of 446 students interviewed believed that race and racism is an important issue. If we want to construct a 90% confidence interval for the true proportion of students who believe that race and racism is an important issue, what would be the margin of error? (express the result in percentage and round to the nearest tenth). QUESTION 3 In a random sample of 736 cows in a country, 220 cows tested positive for brucellosis. Calculate the 99% upper level for the true proportion of brucellosis in the population of cows. (express the result in decimals and round to the nearest ten thousandths. i.e. 4 decimal places). QUESTION 4 A random sample of 65 students at the University has a body weight of 84Kg. It is known from past history that the o- (population variance) of the body weight is 124 Kg 1. Calculate the upper 95% confidence interval for the population mean. (round to the nearest tenth; place your answer in the box below) 2. If we change the confidence level from 95% to 80% when constructing a Cl for the population mean, we can expect the size of the interval to increase/decrease; place your answer right beside the confidence interval without a space and without brackets) Full credit will be given if 1 and 2 are correct. Check your spelling!

Answer 1

solution - 2) ariven that in a survey performed at the university it was found that 46 46 out out of 446 students interviewed believed that race and racism is an important issue. sample proportion of students who believe that race and racism is an important issue p = 46 ५५6 = 0.103 The gol. confidence interval for true proportion" (b) of students who believed that race and racism is an important issue is given by ( § - Zo.us SEC ) , ſ + Z0.05 S.ECH) (since a=0.10 a = 0.05. Hence Zoos is used) Here S.E(6) = 16 (1-96) n 0-103 C1-0.103) V 446 10.103 X 0.897 - 446 - 10.00@21 = 0.046 Hence confidence interval for ļ is (0.103 – 1.96 X 0.046, 0.103+1-96 x 0.046) = (0.01284, 0.19316) = (0.01, 0.13) ĭ 20.05 = 1.96 from table y Margin of error = critical value x SE of the samples = 1.96 X 0.046 = 0.99 = 0.09 X 100%.

Given that in a random sample of 736 cows in a country , 220 cows tested positive for brucellosis. sample proportion of cours tested positive for brucellosis (p)= 220 736 = 0.2989 ggt upper level for the true proportion of brucellosis in the population given by table þ + zool S.Ele Y Zooi = 2.33 from SECB) - JBC1-6) n 0.2989 C1-0-2989) 736 V 0.00028 0.0167 = 1) = 0.2989 given that a random sample of 65 students at the university has a body weight of 84 kg 02 ( population variance) = 124 kg upper 951. confidence interval for the population mean (o, + 20.05 = (0,84 +1.96 v124 - (0,84 +1-96 11:14 - (0, 87.65) V 65 S-

If we change the confidence level from 95% to 80% when constructing a c. I for the population mean, we can expect the size of the interval to decrease ... since a increases from 0.05 to 0.2 which results in the decrease of confidence interval for population mean.