Question

# C) A random survey of 436 workers showed that 192 of them said that it is...

C) A random survey of 436 workers showed that 192 of them said that it is unethical to monitor employee email. When 121 senior-level bosses are surveyed, 40 say that it is unethical to monitor employee email. Use a 0.05 confidence level to test that claim that the proportion of workers is higher than the proportion of senior-level bosses who believe that it is unethical to monitor employee email.25. Use this to construct a 90% C. I. Interpret your results. (carry 3 decimal points.)

Given,

sample proportion p1^ = x1/n1 = 192/436 = 0.44

p2^ = x2/n2 = 40/121 = 0.33

p^ = (x1 + x2)/(n1 + n2) = (192 + 40)/(436 + 121) = 0.417

q^ = 1 - p^ = 1 - 0.417 = 0.583

Ho : p1-p2 <= 0

Ha : p1 -p2 > 0

Here it is a upper tailed test

alpha = 0.05

Critical value = 1.645 [since from z table]

Rejection region:

Reject Ho, if z > 1.645

consider,test statistic z = (p1^ - p2^)/sqrt(pq(1/n1 + 1/n2))

substitute values

= (0.44 - 0.33)/sqrt(0.417*0.583(1/436 + 1/121))

z = 2.17

Here we observe that, test statistic > critical value, so we reject Ho.

So here there is a sufficient evidence.

Now consider,

Here at 90% CI, z value is 1.645

90% CI = (p1^ - p2^) +/- z*sqrt(pq(1/n1 + 1/n2))

substitute values

= (0.44 - 0.33) +/- 1.645*sqrt(0.417*0.583(1/436 + 1/121))

= 0.11 +/- 0.083

= (0.027 , 0.193)

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