In a randomly selected sample of 100 students in a large college, 14 were left-h

Question

In a randomly selected sample of 100 students in a large college, 14 were left-handed. a. Does this provide strong evidence that more than 10% of college students in America are left-handed? Test using alpha = 0.05 b. In part (a), what type of error might you have committed? c. What is the type II error rate of the test conducted in part (a) if the true proportion of left-handers is 0.13 and a sample size of 100 is used? d. How many college students are needed to test that the power 1 - beta (0.13) is 80% for the test of part (a)?

Explanation / Answer

p = 0.14

Null HYpothesis : p = p0 =0.10

Alternative hypothesis : p > p0 = 0.10

(a) Test Statistic

Z = (p - po) / sqrt [ p0 (1-p0))/n] = (0.14 - 0.10)/ sqrt [ 0.10 * 0.90/100] = 0.04/ 0.03 = 1.33

for alpha = 0.05 , Zcritical = 1.645 for one tailed test

so we can not reject the null hypothesis so there is no strong evidence that more than 10% of college students in America are left handed.

(b) In part(a), I might have committed type II error as i have not rejected the null hypothesis so it is the only type of error i can commit.

(c) True proportion of left handers is 0.13 and sample size = 100

so Type II error will be committed when we will not reject the null hypothesis even if it is false.

so Pr( Type II error) = Pr ( p <= 0.10 + 1.645 * 0.03 ; 0.13; 0.03) = Pr( p<= 0.14935, 0.13; 0.03)

Z = ( 0.14935 - 0.13)/ 0.03 = 0.645

so Pr(Type II error) = 0.74

(d) HEre power = 80%

so that means probability of type II error must be 20% or 0.2

soPr( p <=0.10 + 1.645 * sqrt(0.1 *0.9/n) ; 0.13 ; sqrt(0.1 *0.9/n)) = 0.2

Z - value for probability equal to 0.2 is -2.055

so Z= -2.055

(0.10 +1.645 * (0.09/n) - 0.13)/ (0.09/n) = -2.055

(1.645 * (0.09/n) - 0.03)/ (0.09/n) = -2.055

0.03 - 1.645 * (0.09/n) = 2.0555 (0.09/n)

0.03 = 3.7 * (0.09/n)

123.33 = n/0.09

n = 1369

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