In: Statistics and Probability
Suppose that you are the director of table game operations at a large casino, known especially for its poker room. You are interested in determining the average number of hands the casino should expect per hour during peak hours in the poker room. You monitor the number of hands over the next week and given a random sample of 22 hours, you see that on average 175 hands are played per hour, with a standard deviation of 8.63 hands and you calculated a 90% confidence interval to be (171.8, 178.2). Your manager believes the true mean is 172.1. Which of the following is the best conclusion?
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Solution : -
Given that,
Point estimate = sample mean = = 175
sample standard deviation = s = 8.63
sample size = n = 22
Degrees of freedom = df = n - 1 = 22 - 1 = 21
At 90% confidence level
= 1 - 90%
=1 - 0.90 = 0.10
/2
= 0.05
t/2,df
= t0.05,21 = 1.721
Margin of error = E = t/2,df * (s /n)
= 1.721 * ( 8.63 / 22 )
Margin of error = E = 3.2
The 90% confidence interval estimate of the population mean is,
- E < < + E
175 - 3.2 < < 175 + 3.2
171.8 < < 178.2
( 171.8 , 178.2 )
The 90% confidence interval estimate of the population mean is : - ( 171.8 , 178.2 )
( b )
= 172.1
= 175
s = 8.63
n = 22
The null and alternative hypothesis is ,
H0 : = 172.1
Ha : > 172.1
This is the right tailed test .
Test statistic = T
= ( - ) / s / n
= ( 175 - 172.1 ) / 8.63 / 22
= 1.58
The test statistic = 1.58
P - value = P(Z > 1.58 ) = 1 - P (Z < 1.58 )
= 1 - 0.9429
= 0.0571
P-value = 0.0571
= 0.10
P-value <
Reject the null hypothesis .
There is sufficient evidence to the test claim .
Answer : - We are 90% confident that the average number of hands played per hour is greater than 172.1
Correct option : - ( 2 )