Let f(x) =

Mathematics

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Oxana [17]2 years ago

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Answer:

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lidiya [134]

<h2>Answer with explanation:</h2>

Given : The proportion of New Zealanders consume five or more servings of soft drinks per week : $\hat{p}= 0.16$

a) The number of survey respondents reported that they consume five or more servings of soft drinks per week = 2006

b) Confidence interval for population proportion (p) :

$\hat{p}\pm z^*\sqrt{\dfrac{\hat{p}(1-\hat{p})}{n}}$

, where $\hat{p}$ = Sample proportion.

n= Sample size.

z* = Critical value.

For n= 2006 , $\hat{p}= 0.16$ and critical value for 95% confidence interval : z* = 1.96

Then , the required confidence interval will be :

$0.16\pm (1.96)\sqrt{\dfrac{ 0.16(1- 0.16)}{2006}}$

$0.16\pm (1.96)\sqrt{ 0.000066999002991}$

$0.16\pm (1.96)(0.00818529186963)$

$0.16\pm 0.0160431720645$

$(0.16-0.0160431720645,\ 0.16+0.0160431720645)$

$(0.143956827935,\ 0.176043172064)\approx(0.144,\ 0.176)$

i.e. A 95% confidence interval for the proportion of New Zealanders who report that they consume five or more servings of soft drinks per week. = $(0.144,\ 0.176)$