Question 1170231: Among 10,000 school children, 500 of them are randomly selected for a survey on nutritional status among school age children. Of these 500 school children, 91 are found to be moderately and severely underweight children. At 5% level of significance, is their evidence that the proportion of moderately and severely underweight children is 20%?
Answer by Boreal(15235)   (Show Source):
You can put this solution on YOUR website! The point estimate is 91/500=0.182. Will not use population correction factor given 5% of the population is known.
Ho: p=0.20
Ha: p NE 0.20
alpha=0.05 p{reject Ho|Ho true}
Critical value |z|>1.96
test stat z=(p hat-p)/sqrt(p*(1-p)/n)
=(0.0182-0.20)/sqrt(0.2*0.8/500)
=-0.0182/0.0179
z=-1.01
fail to reject Ho.
There is insufficient evidence to conclude that the proportion is not 20%, so there is evidence that the proportion of these children is 20%.
p-value=0.31
With the population correction factor, the std error would be affected by the sqrt (N-n)/(n-1) or sqrt (9500/9999) or 0.9948, and this would not increase the z-value enough to significantly change the result.
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