Question

In: Statistics and Probability

A National Health Survey of 16, 000 children between 2 and 18 found that 40% of...

A National Health Survey of 16, 000 children between 2 and 18 found that 40% of the daily calory intake was from sugar

a) What is our SE measure for this data?



b) What is the value for a 99% confidence interval for p?

c) Write out the confidence interval.



Solutions

Expert Solution

Sol:

a) What is our SE measure for this data?

p^=0.40

SE=sqrt(p^*(1-p)^/n)

=sqrt(0.40*(1-0.40)/16000)

=0.003872983

b) What is the value for a 99% confidence interval for p?

zcrit for 99%=2.576

99% confidence interval for p is

p^-Z*SE,p^+Z*SE

0.40-2.576*0.003872983,0.40+2.576*0.003872983

0.3900232,0.4099768

c) Write out the confidence interval.

0.3900232<p<0.4099768

99% confidence interval for p is

[0.3900232,0.4099768]

orchestra answered 2 years ago
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