Third question stats hw math?

3. email and privacy. A survey of 436 workers showed that 192 of the said that it was seriously unethical to monitor employee e-mail. when 121 senior-level bosses were surveyed 40 said that it was seriosly unethical to monitor employee email ( based on data from a gallup poll) Use a 0.05 significance level to test the claim that for those saying that monitoring email is seriously unethical the proportion of employees is greater than the proportion of bosses.





Ho:


Ha:


mean: .05


Rejection Region:


Calc=2.17


Formula is:





Rejection Ho or cannot Reject Ho ( which one)





Is the proportion of workers greater than the proportion of bosses?





P value?

Third question stats hw math?
These surveys have yes/no (two categorical) answers, so


it's an example of a binomial experiment.


These binomial distributions, with fairly large samples, can be approximized by a normal distribution N(np , sqrt(npq)), with these surveys we have


workers answers : p= 192/436 = 0.44, q = 1-p = 0.56


X ~ N(192,sqrt(0.44*0.56*436) = N ( 192, 107.4 )


Bosses answers : p = 40/121 = 0.33 ; q = 1-p = 0.67


Y ~ N(40, 0.33*0.67*120) = N (40, 26.75).


We can now test on equal means with the Student-t test:


T1 = Sqrt ( (n1*n2*(n1+n2-2)/(n1+n2))


T2 = (m_x - m_y)/ sqrt((n1-1)*s²_x + (n2-1)*s²_y )


T0 = T1*T2 = 2.17 (you say)


The solution c of the equation p(T%26gt;c)=0.95 can be read from the table of the Student-t-distribution with (n1+n2-2)=555 d.f.


Giving c = 1.65.


Because T0 = 2.17 %26gt; 1.96 we reject the null hypothesis of equal means in favour of a significant difference of opinion.

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