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.

(1) Is there evidence that the population mean loss amount is greater than $45 at the 95% confidence level? Explicitly state your null and alternative hypotheses.

(2) What is the p–value = _____

Answer 1

Solution :

Given that,

Population mean = $\mu$ = 45

Sample mean = $\bar x$ = 55

Sample standard deviation = s = 40

Sample size = n = 60

Level of significance = $\alpha$ = 0.05

This a right (One) tailed test.

The null and alternative hypothesis is,  

Ho: $\mu$ $\leq$ 45

Ha: $\mu$ 45

The test statistics,

t = ( $\bar x$ - $\mu$ )/ (s/$\sqrt{n}$)

= ( 55 - 45 ) / ( 40 / $\sqrt{}$ 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.

1)

Conclusion :

It is concluded that the null hypothesis Ho is rejected. Therefore, there is enough evidence to claim that the population mean loss amount is greater than $45 at the 95% confidence level.

The 95% confidence interval is 44.667 < $\mu$ < 65.333

2)

p–value = 0.0288