Question

In: Statistics and Probability

Ages of Gamblers The mean age of a sample of 24 people who were playing the...

Ages of Gamblers The mean age of a sample of 24 people who were playing the slot machines is 49.1 years, and the standard deviation is 6.8 years. The mean age of a sample of 31 people who were playing roulette is 55.2 with a standard deviation of 3.2 years. Can it be concluded at =α0.01 that the mean age of those playing the slot machines is less than those playing roulette? Use μ1 for the mean age of those playing slot machines. Assume the variables are normally distributed and the variances are unequal.

Solutions

Expert Solution

T-test for two Means – Unknown Population Standard Deviations - Unequal Variance

The following information about the samples has been provided:
a. Sample Means : Xˉ1​=49.1 and Xˉ2​=55.2
b. Sample Standard deviation: s1=6.8 and s2=3.2
c. Sample size: n1=24 and n2=31

(1) Null and Alternative Hypotheses
The following null and alternative hypotheses need to be tested:
Ho: μ1 =μ2
Ha: μ1 <μ2
This corresponds to a Left-tailed test, for which a t-test for two population means, with two independent samples, with unknown population standard deviations will be used.

(2) The degrees of freedom
Assuming that the population variances are unequal, the degrees of freedom are given by


(3a) Critical Value
Based on the information provided, the significance level is α=0.01, and the degree of freedom is 30.8671. Therefore the critical value for this Left-tailed test is tc​=-2.4573. This can be found by either using excel or the t distribution table.

(3b) Rejection Region
The rejection region for this Left-tailed test is t<-2.4573

(4)Test Statistics
The t-statistic is computed as follows:


(5) The p-value
The p-value is the probability of obtaining sample results as extreme or more extreme than the sample results obtained, under the assumption that the null hypothesis is true. In this case,
the p-value is 0.0002

(6) The Decision about the null hypothesis
(a) Using traditional method
Since it is observed that t=-4.0604 < tc​=-2.4573, it is then concluded that the null hypothesis is rejected.

(b) Using p-value method
Using the P-value approach: The p-value is p=0.0002, and since p=0.0002≤0.01, it is concluded that the null hypothesis is rejected.

(7) Conclusion
It is concluded that the null hypothesis Ho is rejected. Therefore, there is enough evidence to claim that the population mean μ1​ is less than μ2, at the 0.01 significance level.


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