In a recent research study concerning nutrition, it was reported that only 2% of
ID: 3221112 • Letter: I
Question
Explanation / Answer
a. n = 500 , x = 16 and p0= 0.02 and p = 16/500 = 0.032
b. non skewness criteria
np >= 10 => 500 * 0.032 = 16>= 10
and n(1-p) >= 10 => 500* ( 1 - 0.032) = 484 >=10
and np(1-p) >= 10 => 500* 0.032 * (1- 0.032) = 15.488 >=10
c. Null Hypothesis : H0: p <= p0proportion of children who eat recommended number of servings is as same as earlier
Alternative Hypothesis: Ha : p > p0proportion of children who eat recommended number of servings has increased from earlier
Test Statistics :
Z = ( p - po)/ sqrt(p0(1 - p0)/n) = ( 0.032 - 0.02)/ sqrt(0.02 * 0.98/500) = 0.012 / 0.00626 =1.9169
Zcriticialfor right tailed Z - test for alpha = 0.05 => Zcriticial = + 1.65
so here we can see that Z< Zcriticialso we can reject the null hypothesis and we can conclude that there is significant increase in proportion of children who eat recommended number of serving after advertisement campaign.
(d) P - value from Z - table for z = + 1.9169
p - Value = 0.0285
which is less than alpha = 0.05
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