Question

A survey asked college freshman how far away from their hometown is from the college campus. A random sample of 3 students is taken with a mean distance of 100 miles and a standard deviation of Complete parts (a) and (b) below (a) Find the 95% confidence interval for the population mean Lower bound Upper bound Round your answer to one decimal place) (6) Suppose you want the time to be within 3 mies of the population mean Determine the minimum sample sice needed to constructa 95% confidence interval for the population man under the condition Minimum sample size Your awwe must be a whole number)

Answer 1

(a) The formula for calculating the $(1-\alpha)100\%$ confidence interval for the population mean, $\mu$ is given by: $\mathrm{\bar{x}\pm t_{(\frac{\alpha}{2},df=n-1)}\left ( \frac{s}{\sqrt{n}} \right )}$

Given: sample size,
n = 36
; sample mean,
T= 100 miles
; sample standard deviation, $s=30\:miles$

Critical value: For 95% confidence,

a = 1-0.95 = 0.05

$\mathrm{t_{(\frac{\alpha}{2},df=n-1)}=t_{(\frac{0.05}{2},df=36-1=35)}=2.030108\approx \mathbf{2.0301}}$

$\mathbf{Lower\:bound}:\:\: \mathrm{\bar{x}- t_{(\frac{\alpha}{2},df=n-1)}\left ( \frac{s}{\sqrt{n}} \right )}=\mathrm{100-2.0301\left(\frac{30}{\sqrt{36}}\right)=89.8495\approx {\color{Blue} 89.8}}$

$\mathbf{Upper\:bound}:\:\: \mathrm{\bar{x}+ t_{(\frac{\alpha}{2},df=n-1)}\left ( \frac{s}{\sqrt{n}} \right )}=\mathrm{100+2.0301\left(\frac{30}{\sqrt{36}}\right)=110.1505\approx {\color{Blue} 110.2}}$

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(b) The formula for calculating the minimum sample size needed to construct 95% confidence interval is given by: $n=\left ( \frac{z_{\frac{\alpha}{2}}*s}{E} \right )^{2}$

where,

E: margin of error and it is given in the question as $E=3\:miles$

For 95% confidence,
a = 1-0.95 = 0.05
, $z_{\frac{0.05}{2}}=\mathbf{1.96}$

$n=\left ( \frac{z_{\frac{\alpha}{2}}*s}{E} \right )^{2}=\left(\frac{1.96*30}{3}\right)^{2}=384.16\approx \mathbf{385}$

So the Mainimum sample size need is ${\color{Blue} n=385}$