In 2003, the Gerber company sponsored a large random survey of the eating habits

Question

In 2003, the Gerber company sponsored a large random survey of the eating habits of American infants and toddlers. Among the many questions parents were asked was whether their child had eaten fried potatoes on the day before the interview. Of the 697 infants 9 to 11 months old in the study, 7% had eaten fried potatoes that day. (a) Calculate a 90% confidence interval for p: (b) Which of the following assumptions are needed for the interval in (a) to be valid? Select all that apply (Use 3 decimal places) The data are obtained randomly The original population is normally distributed. The sample size is large enough so that the number of successes and the number of failures are both at least 15. The sample size is large enough so that the number of successes and the number of failures are both less than 15.

Explanation / Answer

Q1.
given that,
possibile chances (x)=48.79
sample size(n)=697
success rate ( p )= x/n = 0.07
I.
sample proportion = 0.07
standard error = Sqrt ( (0.07*0.93) /697) )
= 0.0097
II.
margin of error = Z a/2 * (stanadard error)
where,
Za/2 = Z-table value
level of significance, = 0.1
from standard normal table, two tailed z /2 =1.645
margin of error = 1.645 * 0.0097
= 0.0159
III.
CI = [ p ± margin of error ]
confidence interval = [0.07 ± 0.0159]
= [ 0.0541 , 0.0859]
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DIRECT METHOD
given that,
possibile chances (x)=48.79
sample size(n)=697
success rate ( p )= x/n = 0.07
CI = confidence interval
confidence interval = [ 0.07 ± 1.645 * Sqrt ( (0.07*0.93) /697) ) ]
= [0.07 - 1.645 * Sqrt ( (0.07*0.93) /697) , 0.07 + 1.645 * Sqrt ( (0.07*0.93) /697) ]
= [0.0541 , 0.0859]
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interpretations:
1. We are 90% sure that the interval [ 0.0541 , 0.0859] contains the true population proportion
2. If a large number of samples are collected, and a confidence interval is created
for each sample, 90% of these intervals will contains the true population proportion

Q2.
Option A,B,C

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