Question

20 multiple choice 4 points
construct the requested confidence interval.
a random sample of 10 parking meters in a beach community showed the following incomes for a day. assume the incomes are normally distributed.
$3.60 $4.50 $2.80 $6.30 $2.60 $5.20 $6.75 $4.25 $8.00 $3.00
construct a 95% confidence interval for the true mean.
hint: s = 1.83
($2.11, $5.34)
($1.35, $2.85)
($4.81, $6.31)
($3.39, $6.01)

Explanation:

Step1: Calculate the sample mean

First, find the sum of the data:

$$3.60 + 4.50+2.80 + 6.30+2.60 + 5.20+6.75 + 4.25+8.00 + 3.00=47$$

The sample mean \(\bar{x}=\frac{47}{10}=4.7\)

Step2: Determine the critical value

Since the sample size \(n = 10\), the degrees of freedom \(df=n - 1=9\). For a 95% confidence interval, the significance level \(\alpha=1 - 0.95 = 0.05\), and \(\frac{\alpha}{2}=0.025\). Looking up in the \(t\) - distribution table, \(t_{\frac{\alpha}{2},df}=t_{0.025,9}=2.262\)

Step3: Calculate the margin of error

The margin of error \(E=t_{\frac{\alpha}{2}}\frac{s}{\sqrt{n}}\), where \(s = 1.83\) and \(n = 10\)

$$E=2.262\times\frac{1.83}{\sqrt{10}}\approx2.262\times0.578\approx1.317$$

Step4: Construct the confidence interval

The confidence interval is \(\bar{x}-E<\mu<\bar{x}+E\)

$$4.7- 1.317<\mu<4.7 + 1.317$$

$$3.383<\mu<6.017\approx(3.39,6.01)$$

Answer:

\((\$3.39,\$6.01)\)