Question

A researcher reports that the average income of dentists in metro manila is Php42000 per month. A sample of 25 dentists has a mean salary of Php43800. At a level of significance or 0.05, test claim that dentists earn more than Php42000 per month. The standard deviation is Php5400.

What should be the null hypothesis?

and alternative hypothesis? What is the level of significance? What is the test statistic (is it t-test or z-test)? What type of test significance should be used (One-Tailed or Two-tailed?)

What is the decision rule?

Answer 1

Given thta, population standard deviation = 5400

sample size ( n ) = 25

sample mean = 43800

i) The null and the alternative hypotheses are,

ii) Level of significance = 0.05

iii) Test statistic is,

Test statistic = Z = 1.67

iv) This test is one-tailed ( right-tailed) test.

v) Critical value at 0.05 level of significance is, Z^* = 1.645

Decision Rule:

Reject the null hypothesis ( H0 ), if test statistic > 1.645

vi) Decision: since, test statistic = 1.67 > 1.645

=> We reject the null hypothesis ( H0 ).