Question

1. State the Null and Alternative
2. Find The Standard error and Test statistics (show work for credit ) include the decoding.
3. Draw the standard normal Distribution and decide how likely is the test statistics.
4. Decide to reject or accept the Null hypothesis ~ Use the Test statistics and or the p-value. Show or describe where did you get the p-value if use it.
5. Conclusion

It is claimed that 6% of the population in a specific village suffer from a certain rare eye disorder. However, the doctor in the village believes that the true proportion of sufferers is actually lower than 6%. He randomly tests 500 people from the village and finds that 22 of them have the eye disorder. Is there significant evidence that the doctor right?

Answer 1

Ho:p=0.06

Ha:p<0.06

standard error=sqrt(p*(1-p)/n)

=sqrt(0.06*(1-0.06)/500)

standard error= 0.01062073

Test statistic,z=p^-p/sqrt(p*(1-p)/n)

p^=x/n=22/500= 0.044

z=p^-p/sqrt(p*(1-p)/n

=diff in proportions/standard error

=(0.044-0.06)/0.01062073

=-1.506488

p value in excel

=NORM.S.DIST(-1.506488,TRUE)

p value=0.065970973

p>0.05

Fail to reject ho

Conclusion:There is no suffcient statisitcal evidence at 5% level of significance to support the claim that  

the true proportion of sufferers is actually lower than 6%.