3. A p lan for an executive travelers\' club has been developed by an airline on

Question

3. A p lan for an executive travelers' club has been developed by an airline on the premise that 5% of its current customers would qualify for a membership. A random (a) Find a 99% confidence interval for the true proportion of customers who would (h) Test the hypothesis that the true proportion is more than 5%. Find the P-value customers yielded 40 who would qualify. qualify and use a o .05 significance level. Be sure to state the null and alternative hypotheses, come to a conclusion, and interpret your result.

Explanation / Answer

QUestion 5.

Probabilty that a random person will qualify for for membership = 0.05

(a) Sample proportion p^ = 40/500 = 0.08

sample size N = 500

99% confidence interval for true porportions of customers who would qualify = p^ +- z99% sqrt[p^ * (1-p^)/N]

= 0.08 +- 2.575 * sqrt [ 0.08 * 0.92 / 500]

= 0.08 +- 2.575 * 0.0121

= ( 0.0488, 0.1112)

(b) H0 : p = 0.05

Ha : p > 0.05

where p is the proportion of current employees qualified for membership

standard error of the proportion = sqrt [0.05 * 0.95/ 500] = 0.0097

confidence level alpha = 0.05

Test statistic

Z = (0.08 - 0.05)/ 0.0097=3.09

P- value = Pr(Z > 3.09) = 0.0010 < 0.05

so we shall reject the null hypothesis and can claim that true proportion is more than 5%.

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