Question

In: Statistics and Probability

Suppose for a paper in Professor Hughes' class, you collected data on campaign expenditures by incumbents...

Suppose for a paper in Professor Hughes' class, you collected data on campaign expenditures by
incumbents and challengers in state legislative elections where the race was competitive (i.e., the election
winner received less than 60% of the vote). The mean of campaign expenditures (measured in $1,000s) by
the 51 challengers in these races is 156.0 with a standard deviation of 24.0, while the mean of campaign
expenditures by the 51 incumbents is 162.5 with a standard deviation of 18.0. Treating these as
independent random samples of all possible challengers and incumbents in competitive state legislative
races, test the null hypothesis that population mean of challenger campaign expenditures is greater than
the population mean of incumbent campaign expenditures. Can you reject this null hypothesis at the 5%
significance level? [12 points]

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​=156 and Xˉ2​=162.5
b. Sample Standard deviation: s1=24 and s2=18
c. Sample size: n1=51 and n2=51

(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 Right-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=0.05, and the degree of freedom is 92.73. Therefore the critical value for this Right-tailed test is tc​=1.6616. This can be found by either using excel or the t distribution table.

(3b) Rejection Region
The rejection region for this Right-tailed test is t>1.6616

(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 0.9374

(6) The Decision about the null hypothesis
(a) Using traditional method
Since it is observed that t=-1.5473 < tc​=1.6616, it is then concluded that the null hypothesis is Not rejected.

(b) Using p-value method
Using the P-value approach: The p-value is p=0.9374, and since p=0.9374>0.05, it is concluded that the null hypothesis is Not rejected.

(7) Conclusion
It is concluded that the null hypothesis Ho is Not rejected. Therefore, there is Not enough evidence to claim that the population mean μ1​ is greater than μ2, at the 0.05 significance level.

Let me know in the comments if anything is not clear. I will reply ASAP! Please do upvote if satisfied!


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