Please explain the solution to this problem. Individuals filing federal income tax returns prior to March 31 received an average refund of $1056. Consider the population of "last-minute" filers who mail their tax return during the last five days of the income tax period (typically April 10 to April 15).
a. A researcher suggests that a reason individuals wait until the last five days is that on average these individuals receive lower refunds than do early filers. Develop appropriate hypotheses such that rejection of Ho will support the researchers contention.
b. For a sample of 400 individuals who filed a tax return between April 10 and 15, the sample mean refund was $910. Based on prior experience, a population standard deviation of $1600 may be assumed. What is the p-value?
c. At alpha = .05, what is the conclusion?
D. Repeat the preceding hypothesis test using the critical value approach.
a. The null hypothesis (Ho) is that there is no difference in average refunds between "last-minute" filers and early filers. The alternative hypothesis (Ha) is that the average refund for "last-minute" filers is lower than that of early filers.
Ho: μ1 = μ2 (where μ1 is the average refund for early filers and μ2 is the average refund for "last-minute" filers)
Ha: μ1 > μ2 (average refund for "last-minute" filers is lower than for early filers)
b. We can use a one-sample t-test to test the hypothesis. The formula for the t-statistic is:
t = (sample mean - population mean) / (population standard deviation / √sample size)
In this case, the sample mean refund is $910, the population standard deviation is $1600, and the sample size is 400. We can calculate the t-statistic using these values.
t = (910 - 1056) / (1600 / √400)
t = -146 / (1600 / 20)
t = -146 / 80
t = -1.825
To find the p-value, we need to compare the t-statistic to the t-distribution with degrees of freedom (df) equal to the sample size minus 1. In this case, df = 400 - 1 = 399. We can use a statistical software or a t-table to find the p-value corresponding to a t-statistic of -1.825.
c. At alpha = 0.05, if the p-value is less than 0.05, we reject the null hypothesis. If the p-value is greater than or equal to 0.05, we fail to reject the null hypothesis.
In this case, we compare the p-value obtained from step b with the significance level (alpha = 0.05). If the p-value is less than 0.05, we reject the null hypothesis. If the p-value is greater than or equal to 0.05, we fail to reject the null hypothesis.
d. The critical value approach involves comparing the test statistic (t-statistic) to the critical value obtained from the t-distribution.
Using a one-tailed test at alpha = 0.05 and df = 399 (from step b), we can find the critical value corresponding to a significance level of 0.05.
We compare the absolute value of the test statistic (-1.825) to the critical value obtained from the t-table. If the absolute value of the test statistic is greater than the critical value, we reject the null hypothesis. If it is less than or equal to the critical value, we fail to reject the null hypothesis.
By using the critical value approach, we can make a conclusion similar to that in part c.
Note: The exact critical value and p-value may vary depending on the specific t-distribution table or software tool used for calculations.