91Ó°ÊÓ

The formula used: The confidence interval$\left(\stackrel{¯}{x}-t_{\frac{a}{2}}×\frac{s}{\sqrt{n}},\stackrel{¯}{x}+t_{\frac{a}{2}}×\frac{s}{\sqrt{n}}\right)$

Let $\mathrm{μ}$be the average cost of a day at an American amusement park for a family of four.

Population s.d., $\mathrm{σ}$is unknown.

Here the sample size $n=25$

We will utilise the $t-$interval approach to identify a$95\%$Cl of $\mathrm{μ}$as population s.d., $\mathrm{σ}$is unknown.

The $100(1-\mathrm{Î\pm })\%\mathrm{CI}$of population mean $\mathrm{μ}$is given by the $t-$interval producer.

$\left(\stackrel{¯}{x}-t_{\frac{a}{2}}×\frac{s}{\sqrt{n}},\stackrel{¯}{x}+t_{\frac{a}{2}}×\frac{s}{\sqrt{n}}\right)$

Where $t_{\frac{a}{2}}$represents the $t-$value with area $\frac{\mathrm{Î\pm }}{2}$to its right and $df=n-1$and$n$ represents the sample size.

Here confidence level$=95\%=100×0.95\%$

$$\begin{gathered} ∴1-\mathrm{Î\pm }=0.95 \\ ⇒\mathrm{Î\pm }=1-0.95 \\ ⇒\mathrm{Î\pm }=0.05 \\ ⇒\frac{\mathrm{Î\pm }}{2}=0.025 \\ df=n-1 \\ =25-1=24 \\ ∴\text{For}df=24,t_{\frac{\mathrm{Î\pm }}{2}}=t_{0.02}=2.064 \end{gathered}$$

Given that the sample mean $\stackrel{¯}{x}=193.32$

Sample S.D $s=26.73$

$$\begin{gathered} ∴95\%\mathrm{Cl}\text{of}\mathrm{μ} \\ =\left(\stackrel{¯}{x}-t_{0.025}\frac{s}{\sqrt{n}},\stackrel{¯}{x}=t_{0.025}\frac{s}{\sqrt{n}}\right) \\ =\left(193.32-2.064×\frac{26.73}{\sqrt{25}},193.32+2.064×\frac{26.73}{\sqrt{25}}\right) \\ =(193.32-11.03,193.32+11.03) \\ =(182.29,204.35) \end{gathered}$$

i.e., we are $95\%$certain that the average cost of a day at an American amusement park for a family of four, $\mathrm{μ}$, is between $\$182.29$ and $\$204.35$