Question

The table below shows which hand is favored by each person from a sample of 100 people (50 male and 50 female). Use the information in the table to answer the following questions. Left-Handed Right-Handed Male 15 35 Female 10 40 Determine the probability that a randomly selected person from the sample will be female or left-handed. Number | % (Round to the nearest whole percentage) Compute a 95% confidence interval for the true proportion of the population that is left-handed. Number % to | Number | % Compute a 95% confidence interval for the true proportion of the population that is right-handed. Number % to | Number | %

Answer 1

A random sample of 100 people (50 males and 50 females) is selected and they are being asked if they use their left hand more or right hand. Define two events A and B as follows: A: The person is female B: The person is left-handed. Number of females = 50 Number of left-handers = 25 (15 + 10) Number of female left-handers = 10 The probability that a randomly selected person is a female or left-handed is calculated as shown below: B ) P(AUB)=P(A) +P(B)-P(A 50 25 10 100 100 100 50+25-10 100 65 100 = 0.65 Thus, the probability that a randomly selected person is a female or left-handed is 0.65 = 65%. Number of people who are left-handed = 25 So, the sample proportion of people who are left-handed (PL)= =0.25 The 95% confidence interval for the true proportion of population that is left-handed is calculated by the following formula: CI = P, 53,/(1P) Vn In the above formula, -, represents the two-tailed critical value of the z-test statistic at 5% level of significance, which is obtained using MS-Excel function as shown in the screenshot below: ENORM.S.INV(0.025)

The above function yields a critical value of 1.96. Substitute all the values in the above formula to compute the confidence interval as shown in the calculations below: CI = 0.25 +1.96, 10.25x(1-0.25) V 100 10.25x(1-0.25) = 0.25 +1.96 V 100 = 0.25-1.96 0.25 0.75 =0.2311V100 = 0.25+(1.96x0.0433) = 0.25 +0.084868 =(0.25-0.084868.0.25 +0.084868) =(0.165132,0.334868) = (0.1651,0.3349) Therefore, the 95% confidence interval for the true proportion of population that is left- handed is (0.1651, 0.3349) = (16%, 33%) Number of people who are right-handed = 75 So, the sample proportion of people who are right-handed (PR)= =0.75 The 95% confidence interval for the true proportion of population that is right-handed is calculated by the following formula: CI = P+- V PR(1-P) n In the above formula, represents the two-tailed critical value of the z-test statistic at 5% level of significance, which is obtained using MS-Excel function as shown in the screenshot below: =NORM.S.INV(0.025) The above function yields a critical value of 1.96. Substitute all the values in the above formula to compute the confidence interval as shown in the calculations below:

0.75x(1-0.75) CI = 0.75-1.96, V 100 10.75x(1-0.75) = 0.75-1.96 100 10.75 0.25 = 0.75+1.96, 100 = 0.75+(1.96x0.0433) = 0.750.084868 =(0.75 -0.084868.0.75 +0.084868) =(0.665132,0.834868) =(0.6651.0.8349) Therefore, the 95% confidence interval for the true proportion of population that is left- handed is (0.6651, 0.8349) = (67%,83%).