Question

For all hypothesis tests, use both the critical value/rejection region and p-value (separate and label each method) and show each of the 5 steps explicitly.

A. The internal revenue service believes that the number of business executives who default on their tasks is larger than the proportion of 'blue collar' workers. Out of 150 business executives, 24 default on their taxes. For a sample of 75 'blue collar' workers has only 8 that default on their taxes. Test the hypothesis that the proportion of business executives who default on their taxes is larger than the proportion of 'blue collar' workers at alpha = .01.

Round all of the following to 3 decimal places: p hat 1, q hat 1, p hat 2, and q hat 2, p hat and q hat. Round test statistic to 2 decimal places.

B. Find a 99% Confidence Interval for the difference of the two proportions.

Answer 1

Let sample one is sample of business executive and sample 2 is sample of blue collar workers.
So we have here, n1 = 150, x1= 24, n2 = 75, x2 = 8, =0.01

,

,

Claim : The proportion of business executives who default on their taxes is larger than the proportion of 'blue collar' workers.

The hypothesis are :

H0 : P1 = P2 v/s H1: P1 > P2

The test statistic is,

= 1.08

p value = p ( z > 1.08 )

= 1 - p ( z 1.08 )

= 1 - 0.8599 --------------( using excel formula " =norm.s.dist(1.08,1) " )

= 0.1539

Here p value > ( 0.01 ) .

Hence we failed to reject null hypothesis.

The critical value is,

Here calculated value of z < critical value of z. Hence we failed to reject null hypothesis.

Conclusion :

There is no sufficient evidence to conclude that the proportion of business executives who default on their taxes is larger than the proportion of 'blue collar' workers.

B. The 99% confidence interval is given by,

{ ( - ) -E , ( - ) + E }

Where,

c = 0.99,

Zc =

Hence margin of error is,

= 0.05

Hence the 99% confidence interval is given by,

{ ( 0.160 -0.107 ) - 0.05 , ( 0.160 - 0.107 ) + 0.05 }

( 0.003, 0.103 )