In a sample of 550 students, 23% of them will plan transferring in next semester

Question

In a sample of 550 students, 23% of them will plan transferring in next semester.

a. Test the claim that less than 25% of all students will plan transferring in next semester if = 0.03

b. Construct a 92% confidence interval of the proportion of all students who will plan transferring in next semester.

c. If the researchers want the error of estimating the proportion of all students who will plan transferring in next semester is no more than 2.5 percentage points, find the sample size needed if the C-level is 90%

Explanation / Answer

Sample size = 550

p ^ = 23% = 0.23

Standard error of the sampling = Sqrt [p^ (1-p^) /N] = sqrt [ 0.23 * 0.77 / 550 ] = 0.018

(a) Null Hypothesis : H0 : p >=0.25

Alternative Hypothesis : Ha : p < 0.25

Test Statistic

Z = (p- po )/ sqrt [p0 (1-p0 )/N] = ( 0.25 - 0.23)/ 0.018 = 1.11

Pr ( p <= 0.25 ; 0.23; 0.018) = ?

by Z - value = (0.25 - 0.23)/ 0.018 = 1.11

Pr ( p <= 0.25 ; 0.23; 0.018) = (1.11) where is the cumulative normal distribution function.

so Pr ( p <= 0.25 ; 0.23; 0.018) = 0.8665

so p - value = 1 - 0.8665 = 0.1335

so p- value is not under significance level so we cannot reject the null hypothesis and conclude that not 25% of all students will plan transferring in next semester.

b. 92 % confidence interval of the proportion of all students

= p^ +- Z0.08 se0 = 0.23 + 1.76 * 0.018 = (0.1983, 0.2617)

(c) HEre the confidence interval is 90% and Error of estimating the proportion = 2.5%

so HEre

Z0.05 * se0 = 0.025

se0 = 0.025/ 1.645 = 0.0152

se0 = sqrt [p (1-p)/N] = sqrt [ 0.23 * 0.77/N] = 0.0152

N = 777

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