Answer to Question #162730 in Quantitative Methods for Sunny

Question

It was believed ( from past data) that professionals across a particular city had an average net income was Rs 54000 per month with a standard deviation of Rs12000. A researcher wanted to determine whether the net income has changed over a period of time. He took a survey of 100 professionals and found that the average net income is Rs 56000. Test at 1% level of significance.

Answer 1

We have that

$\mu=54000$

$\sigma=12000$

$n=100$

$\bar x=56000$

$\alpha=0.01$

$H_0:\mu=54000$

$H_a:\mu \ne54000$

The hypothesis test is two-tailed.

The population standard deviation is known and the sample size is large (n≥30) so we use z-test.

The critical value for α=1% is Z_0.01 = ±2.57

The critical region is Z < -2.57 and Z > 2.57

Test statistic:

$$Z_{test}=\frac{\bar x -\mu}{\frac{\sigma}{\sqrt n}}=\frac{56000 -54000}{\frac{12000}{\sqrt {100}}}=1.66$$

Since -2.57 < 1.64 < 2.57 thus the Z_test does not fall in the rejection region we fail to reject the null hypothesis. Therefore there is no sufficient evidence to conclude that the net income has changed over a period of time.