In: Statistics and Probability
A mortgage company performs reviews of its closing documents prior to the closing dates. This review may prompt changes or document additions that should have been in the closing package. Any closing package found with inaccuracies or known to be incomplete are deemed defective and require rework prior to closing. In the prior calendar year, the defective rate for these reviews was 17.99% (783 defective packages out of 4,352 reviewed). During the current year, the year-to-date defective rate is 20.69% (180 of 870). Use the information provided to perform a 2-sample proportion test to determine if there is a difference between the current satisfaction rate and the prior year.
given data are:-
sample size In the prior calendar year () = 4352
sample proportion of defective packages in the prior calendar year () = 17.99% = 0.1799
sample size In the current year () = 870
sample proportion of defective packages in the current year () = 20.69% = 0.2069
pooled proportion be:-
hypothesis:-
the test statistic be:-
p value be:-
[ in any blank cell of excel type =NORMSDIST(-1.8747) press enter]
decision:-
p value = 0.0608 >0.05 (alpha)
so, we fail to reject the null hypothesis.
conclusion:-
there is not sufficient evidence to claim that there is a difference between the current satisfaction rate and the prior year at 0.05 level of significance.
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