Answer to Question #156859 in Statistics and Probability for aira

Question

The mayor of a city suspects that the average family income in his city hall has fallen since the last 

decennial was P4,200.00, a random sample of 50 families is taken. This sample has a mean income of 

P13,420.00 and a standard deviation of P4,360.00 can the mayor conclude that the mean income has fallen? 

Use .05 level of significance

Answer 1

Here we have that

$\mu=4200$

$n = 50$

$\bar x=13420$

$s =4360$

$\alpha = 0.05$

$H_0:\mu=4200$

$H_a:\mu<4200$

The hypothesis test is left-tailed.

The test to be used is t-test for the population mean as the population standard deviation is unknown, the sample is random and greater than 30.

df = n – 1 = 49

The critical value for a=0.05 and 49 degrees of freedom is –1.677.

The critical region is t < –1.677.

Test statistic:

$$t=\frac{\bar x - \mu}{\frac{s}{\sqrt n}}=\frac{13420 - 4200}{\frac{4360}{\sqrt 50}}=14.95$$

Since 14.95 > –1.677 thus t does not fall in the rejection region we fail to reject the null hypothesis. There is no sufficient evidence to conclude that the mean income has fallen.