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In a study conducted by the US department of Health and Human Services, a sample

ID: 3154717 • Letter: I

Question

In a study conducted by the US department of Health and Human Services, a sample of 546 boys aged 6-11 was weighed, and it was determined that 87 of them were overweighted. A sample of 508 girls aged 6-11 was also weighed, and 74 of them were overweighted. Find a 90% confidence level for the difference of overweight proportion for aged 6-11 between boys and girls. Can you conclude that the proportion of overweight boys is higher than the proportion of girls who are overweight? State hypothesis, Compute P-value. What is your conclusion?

Explanation / Answer

a) The 90% c.i: (Ps1-Ps2)+-Zcritical SE(Ps1-Ps2), where Ps1 and Ps2 denote proportion of boy sand girls, aged 6-11, Z correspond to Z critical avalue at alpha=0.10 and SE(Ps1-Ps2) correspond to standard error of the two proportions.

SE(Ps1-Ps2)=sqrt [Ps1(1-Ps1)/N1+Ps2(1-Ps2)/N2]

=[{(87/546(1-87/546)}/546+{74/508(1-74/508)}/508]

=0.0221

90% c.i=(87/546-74/508)+-1.645*0.0221

=-0.022 to 0.050