Q 8.132. - Elementary Statistics

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: $\overline{x}=\backslash (193.32; s=\backslash )26.73$)

Step-by-Step Solution

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 1: 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

Step 2: Concept

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

Step 3: 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 $d f=n-1$ and $n$ represents the sample size.

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

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

Step 4: 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 $\$204.35$