Question

Family Fun? Taking the family to an amusement park has become increasingly costly according to the industry publication Amusement Business, which provides figures on the cost for a family of four to spend the day at one of America's amusement parks. A random sample of $25$families of four that attended amusement parks yielded the following costs, rounded to the nearest dollar.

.Obtain and interpret a $95\%$confidence interval for the mean cost of a family of four to spend the day at an American amusement park. (Note: $\overline{x}=\backslash (193.32;s=\backslash )26.73$)

Short answer

we are $95\%$certain that the average cost of a day at an American amusement park for a family of four, $\mu$, is between $\$182.29$ and $\$204.35$

Step-by-step solution

Given information

156	212	218	189	172
221	175	208	152	184
209	195	207	179	181
202	166	213	221	237
130	217	161	208	220

Concept

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

Calculation

Let $\mu$be the average cost of a day at an American amusement park for a family of four.

Population s.d., $\sigma$is unknown.

Here the sample size $n=25$

We will utilise the $t-$interval approach to identify a$95\%$Cl of $\mu$as population s.d., $\sigma$is unknown.

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

$\left(\overline{x}-t_{\frac{a}{2}}\times \frac{s}{\sqrt{n}},\overline{x}+t_{\frac{a}{2}}\times \frac{s}{\sqrt{n}}\right)$

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

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

$$\begin{gathered} \therefore 1-\alpha =0.95 \\ \Rightarrow \alpha =1-0.95 \\ \Rightarrow \alpha =0.05 \\ \Rightarrow \frac{\alpha }{2}=0.025 \\ df=n-1 \\ =25-1=24 \\ \therefore \text{For}df=24,t_{\frac{\alpha }{2}}=t_{0.02}=2.064 \end{gathered}$$

Calculation

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

Sample S.D $s=26.73$

$$\begin{gathered} \therefore 95\%\mathrm{Cl}\text{of}\mu \\ =\left(\overline{x}-t_{0.025}\frac{s}{\sqrt{n}},\overline{x}=t_{0.025}\frac{s}{\sqrt{n}}\right) \\ =\left(193.32-2.064\times \frac{26.73}{\sqrt{25}},193.32+2.064\times \frac{26.73}{\sqrt{25}}\right) \\ =(193.32-11.03,193.32+11.03) \\ =(182.29,204.35) \end{gathered}$$

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