Question

A betting analyst in Las Vegas wants to study the losses suffered from gamblers at a particular casino to determine whether a particular casino is cheating. In particular, the analyst wants to see if gamblers' average losses exceed $45, which is the average from all other casinos. She selects a random sample of 60 gamblers and finds that the sample mean loss was $55 and the sample standard deviation was $40.

Determine the p–value for your hypothesis test. p–value = _____

Answer 1

Solution :

Given that,

Population mean = = 45

Sample mean = = 55

Sample standard deviation = s = 40

Sample size = n = 60

Level of significance = = 0.05

This a right (One) tailed test.

The null and alternative hypothesis is,  

Ho: 45

Ha: 45

The test statistics,

t = ( - )/ (s/)

= ( 55 - 45 ) / ( 40 / 60)

= 1.936

df = n - 1 = 59

P- Value = 0.0288   

The p-value is p = 0.0288 < 0.05, it is concluded that the null hypothesis is rejected.

Conclusion :

It is concluded that the null hypothesis Ho is rejected. Therefore, there is enough evidence to claim that the population mean μ is exceed $45, at the 0.05 significance level.

p–value = 0.0288