Answer to Question #241095 in Statistics and Probability for darv

Question

A sample survey on the average total yearly expenditure included 150 students of a certain university. The mean total expenditure per student per year for the sample was 3,000 with a standard deviation of 500. How likely is it that the students spend an average of 3500 per year as claimed by a parent at .01 significant level. 

Answer 1

$n=150 \\ \bar{x}=3000 \\ s=500 \\ H_0: \mu = 3500 \\ H_1: \mu ≠3500 \\ α=0.01$

When dealing with a large sample of size n >30 from a population that need not be normal but has a finite variance, we can use the central limit theorem to justify using the test for normal populations. Even when $\sigma^2$ is unknown we can approximate its value with $s^2$ in the computation of the test statistic.

Test-statistic:

$Z= \frac{\bar{x} - \mu}{s/ \sqrt{n}}$

$\bar{x}$ = sample mean

$\mu$ = population mean

s = sample standard deviation

n = sample size

$Z = \frac{3000-3500}{500 / \sqrt{150}} = \frac{-500}{40.85} = -12.24$

Two-tailed test. Reject H₀ if Z≤ -2.575 or Z ≥ 2.575.

$Z = -12.24 < Z_{tab} = -2.575$

We reject H₀.

There is NOT enough evidence, that the students spend an average of 3500 per year as claimed by a parent at 0.01 significant level.