A survey cammissland by a health care group rapartad that 21 of New Zealanders c

Question

A survey cammissland by a health care group rapartad that 21 of New Zealanders consume five ar more serings a soft drinks Per woek h data were Db ained by an online survey af 2 o random y salactad New Zoa anders over 15 years of age. What number of survey respondents reported that they consume tive or more servings of soft drinks per week? (Round your ansiwer to the nearest whole number.) Why was it necessary to round your answer in the previous question? Rounding to the nearast whola numbar was necessary because the question asked for a number Rounding to the nearest whole number was necessary because there were 2,009 people in the survey. Rounding to the nearest whole number was necessary because the question asked about a percent Rounding to the nearast whole number was necessary because the number was bigger than one hundrad Rounding to the nearest whole number was necessery because you can not have fractions of people (b) Find a 95% confidence interval for the proportion of New Zealanders who report that they consume five or more servings of soft drinks per week. (Round your answers to four decimal places.) c) Convert the estimate and your confidence interval to percents. Round your answers to two decimal places.) (d) Discuss reasons why the estimate might be biased. (Select l that apply.) The survey did not include people under 15 years of age ) The survey only included New Zealanders. The survey had 2,009 people. The survey questions may have been biased. O The survey was commissloned by a health care group. The survey respondents might have ed

Explanation / Answer

a.
given that,
possibile chances (x)=421.89 ~ 422
sample size(n)=2000
success rate ( p )= x/n = 0.21

requesting to the nearest number is necessary because the question asked for a number

b.
I.
sample proportion = 0.21
standard error = Sqrt ( (0.21*0.79) /2000) )
= 0.0091
II.
margin of error = Z a/2 * (stanadard error)
where,
Za/2 = Z-table value
level of significance, = 0.05
from standard normal table, two tailed z /2 =1.96
margin of error = 1.96 * 0.0091
= 0.0179
III.
CI = [ p ± margin of error ]
confidence interval = [0.21 ± 0.0179]
= [ 0.1921 , 0.2279]
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DIRECT METHOD
given that,
possibile chances (x)=420
sample size(n)=2000
success rate ( p )= x/n = 0.21
CI = confidence interval
confidence interval = [ 0.21 ± 1.96 * Sqrt ( (0.21*0.79) /2000) ) ]
= [0.21 - 1.96 * Sqrt ( (0.21*0.79) /2000) , 0.21 + 1.96 * Sqrt ( (0.21*0.79) /2000) ]
= [0.1921 , 0.2279]

c.
19.21% to 22.79%

d.
the survey is did n't include people under age 15
survey questions may be biased
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interpretations:
1. We are 95% sure that the interval [ 0.1921 , 0.2279] contains the true population proportion
2. If a large number of samples are collected, and a confidence interval is created
for each sample, 95% of these intervals will contains the true population proportion

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