1.An article in a professional journal reports the z­score for the study’s sampl

Question

1.An article in a professional journal reports the z­score for the study’s sample mean was –1.4. The authors also reported their sample mean was 97 with a sample size of 49 and the population mean was 102. However, they did not give the population standard deviation. What was their population standard deviation?

2.A researcher wanted to compare the self-esteem scores of children of divorced parents to the general population. To do so, she measured self-esteem in 100 children of divorced parents and got an sample mean for self-esteem score of 27. If the population self-esteem average is =30 with a standard deviation of =7, run a full hypothesis test using the z-test to test the hypothesis that children of divorced parents have significantly different levels of self-esteem. Use an alpha of =0.05, two-tailed test.

Explanation / Answer

let population std deviation be x

hence z=(X-mean)/(std deviation/(n)1/2)

-1.4=(97-102)(x/(49)1/2)

x=25

b)here std error =7/(100)1/2 =0.7

hence test stat z=(X-mean)/std error =(27-30)/0.7 =-4.2857

for whcih p value =0.0000

as p value is less then 0.05 hece we reject null hypothesis that mean score of children of divorced parents is 30

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