A) Construct the 98% confidence interval for the true proportion of children who

Question

A) Construct the 98% confidence interval for the true proportion of children who are deficient in vitamin D B) Explain carefully what the interval means C) Explain what 98% confidence means This Question: 1 pt 3of5(°complete) Arecent shady of 3400 children randomly selected found 23% of them defeertnvtann D zjConstructthe98% con dence interval for the true proporton ofchildren who are deficient n vitamnD Round to three decimal places as needed) bi Explain carfully what the interval means OA, we are 96% confident that the rterval contains the true proporton of people descent n stan D B. we are 98% confdent that the nterval contanstho true proporton of thkben delcient in wtamnD OD, wc ao 98% confden tutte percent of people de kent invernn0s23% Atot 90% of random samples or sine 3400 wa pod ice eat contain the of children that ase deficient n vitam popslation proptio confderee intorval bua propo

Explanation / Answer

A)

Sample proportion=p

q=1-p

Sample size=n

Confidence interval=p±z*SE(p)SE(p)=[(pq) / (n)]

ME=z*[(pq) / (n)]

SE(p)=[(pq) / (n)]SE(p)=[(0.23 x 0.77) / (3400)]SE(p)=0.007

z*for a 98% confidence interval=2.33

ME=z*[(pq) / (n)]

ME=2.33 x[(0.23 x 0.77) / (3400)]

ME=0.016

Lower confidence limit=p-z*SE(p)

Lower confidence limit=0.23-2.33(0.007)

Lower confidence limit=0.214

Upper confidence limit=p+z*SE(p)

Upper confidence limit=0.23+2.33(0.007)

Upper confidence limit=0.24698%

confidence interval: (0.214,0.246) or 21.4% to 24.6%

B)

What 98% confidence interval states is that, when you do this experiment by sampling the same data multiple times or estimating the population parameter using sampling, 98 out 100 times the true population parameter will lie within the range of your estimate

C)

We are 98% confident that the value lies in the calculated interval. You don't want to put we are 98% sure that whatever, always state it as 98% confident.

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