Answer to Question #137679 in Microeconomics for Massina Veremaito

Question

Some high school teachers, teaching schools in outer islands demand an increase in meal allowance. According to a recent survey, the average cost of a meal is $6 in Ba. Suppose this figure is based on a sample of 27 different restaurants with a sample standard deviation of $1. Construct a 95% confidence interval for the population mean cost of a fast food meal in Ba. Assume that the cost of fast food meals in Ba is normally distributed.

Answer 1

$\bold {Answer}$

The 95% Confidence interval $=(\$5.60, \space \$6.40)$

$\bold {Solution}$

Since the sample size (n) is less than 30, we use the t-statistic to compute the required confidence interval (CI).

$CI: \space \mu = \bar{x} \pm t_{\alpha/2} \big(\dfrac {s}{\sqrt {n}} \bigl)$

We are provided:

$\bar {x} = \$6$

$s = \$1$

$n = 27$

Degrees of freedom:

$\nu = n -1$

$= 27 -1$

$= 26$

$\alpha = 1 -0.95$

$= 0.05$

$\alpha/2 = \dfrac {0.05}{2}$

$= 0.025$

$t_{0.025} (26) = 2.056$ (shown on the table provided below)

$Thus, \space \mu = \bar {x} \pm t_{\alpha/2} \big(\dfrac {s}{\sqrt {n}} \big)$

$= 6 \pm 2.056 \big (\dfrac {1}{\sqrt {27}} \big)$

$= 6 \pm 0.3956773844$

Lower Limit:

$\mu = 6-0.3956773844$

$= \$5.604322615$

$\approx \$5.60$

Upper Limit:

$\mu = 6 + 0.3956773844$

$= \$6.3956773844$

$\approx \$6.40$

Confidence Interval:

$\mu$ lies in the interval $(\$5.60, \space \$6.40)$

Therefore, the value of the mean cost of a fast food meal in Ba is contained in the interval ($5.60, $6.40) with 95% confidence.