Question

a sample of 20 part - time workers had mean annual earnings of $3120 with a standard deviation of $677. construct a 95% confidence interval for the population mean μ. assume the population has a normal distribution. round to the nearest dollar

a. ($1324, $1567)
b. ($2657, $2891)
c. ($2803, $3437)
d. ($2135, $2587)

Explanation:

Step1: Determine the critical value

Since the sample size \(n = 20\) (small - sample, \(n<30\)) and the population is assumed to be normal, we use the \(t\) - distribution. The degree of freedom \(df=n - 1=20 - 1 = 19\). For a \(95\%\) confidence interval, the significance level \(\alpha=1 - 0.95 = 0.05\), and \(\frac{\alpha}{2}=0.025\). Looking up the \(t\) - value in the \(t\) - distribution table, \(t_{\frac{\alpha}{2},df}=t_{0.025,19}=2.093\).

Step2: Calculate the margin of error

The formula for the margin of error \(E=t_{\frac{\alpha}{2}}\frac{s}{\sqrt{n}}\), where \(s = 677\) (sample standard deviation) and \(n = 20\).

$$
LATEXBLOCK0
$$

Step3: Calculate the confidence interval

The formula for the confidence interval is \(\bar{x}-E<\mu<\bar{x} + E\), where \(\bar{x}=3120\) (sample mean).

$$
LATEXBLOCK1
$$

Rounding to the nearest dollar, the confidence interval is \((2803,3437)\)

Answer:

C. (\$2803, \$3437)