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Given in the question that,

n = $600$students

x =$480$

Compute a$90$% confidence interval for the true percent of students who are against the new legislation, and interpret the confidence interval.

If $\stackrel{鈭�}{p}$is the proportion of observations in a random sample of size n that belongs to a class of interest, an approximate $100$($1$-$伪$)% confidence interval on the proportion p of the population that belongs to this class is

$\hat{p}鈭�z_{\frac{伪}{2}}\sqrt{\frac{\hat{p}(1鈭�\hat{p})}{n}}鈮�p鈮�\hat{p}+z_{\frac{伪}{2}}\sqrt{\frac{\hat{p}(1鈭�\hat{p})}{n}}$ (1)

$\text{where}{z}_{\frac{伪}{2}}\text{is the upper}\frac{伪}{2}\text{percentage point of the standard normal distribution.}$

This method depends on the sufficiency of the normal approximation to the binomial. To be reasonably conservative, this requires that np and n($1$-p) be greater than or equal to $5$. In situations when this approximation is inappropriate, particularly in cases when n is small, other methods must be used.

a) We know surveys n=$600$students and finds that x=$480$are against the new legislation. Then, the point estimate of the proportion of the population p is

$\hat{p}=\frac{x}{n}=\frac{480}{600}=0.8$

and

$np=600脳0.8=480>5$

$n(1鈭�p)=600脳(1鈭�0.8)=120>5$

$\text{For}90\mathrm{\%}\text{two-sided confidence interval,}$

$\frac{伪}{2}=\frac{1鈭�0.90}{2}=0.05$

and

$z_{\frac{伪}{2}}=1.645$

Therefore, from Equations ($1$) and ($2$) a $95$% two-sided CI estimate for the true proportion of people who own tablets is

$0.8鈭�1.645\sqrt{\frac{0.8(1鈭�0.8)}{600}}鈮�p鈮�0.8+1.645\sqrt{\frac{0.8(1鈭�0.8)}{600}}$

which simplifies to

$0.773鈮�p鈮�0.827$

Given in the question that,

n = $600$students

x = $480$

In a sample of $300$students , $68$% said they own an iPod and a smartphone. Compute a$97$% confidence interval for the true percent of students who own an iPod and a smartphone

In a sample of n=$300$students, $68$% said they possess an iPod and a smartphone. Then, the point calculation of the proportion of the population, p is

$\hat{p}=68\mathrm{\%}=0.68$

and

$np=300脳0.68=204>5$

$n(1鈭�p)=300脳(1鈭�0.68)=96>5.$

$\text{For}97\mathrm{\%}\text{two-sided confidence interval,}$

$\frac{伪}{2}=\frac{1鈭�0.97}{2}=0.015$

and

$z_{\frac{伪}{2}}=2.17$

Therefore, from Equation ($1$) a $97$% two-sided Cl for the true percent of students who own an iPod and a smartphone is

$0.68鈭�2.17\sqrt{\frac{0.68(1鈭�0.68)}{300}}鈮�p鈮�0.68+2.17\sqrt{\frac{0.68(1鈭�0.68)}{300}}$

which simplifies to,

$0.622鈮�p鈮�0.738$