It is known that the population mean for the Quantitative section of the GRE is 500, with a standard deviation of 100. In 2006, a sample of 400 students taking the GRE, whose family income was between $70,000 and $80,000, had a quantitative GRE score of 511. Perform a one-tailed hypothesis test to determine whether the group, whose family income was between
$70,000 and $80,000, scored significantly higher on average than the population.
10. What is the value of the test statistic?
a. .11
b. -.11
c. 2.2
d. -2.2
11. What is the p-value for the test statistic?
a. .4562
b. .5438
c. .0139
d. .9861
12. Using the significance level of .05, do the results lead to rejection of the null hypothesis?
Please describe what the results tell us about mean quantitative GRE scores for those with
a family income between $70,000 and $80,000.
a. Yes, reject the null hypothesis. The mean quantitative GRE score for students, whose family income was between $70,000 and $80,000, is not significantly higher than then
the average quantitative GRE score for the population.
b. Yes, reject the null hypothesis. The mean quantitative GRE score for students, whose family income was between $70,000 and $80,000, is significantly higher than then the
average quantitative GRE score for the population.
c. No, do not reject the null hypothesis. The mean quantitative GRE score for students, whose family income was between $70,000 and $80,000, is not significantly higher than
then the average quantitative GRE score for the population.
d. No, do not reject the null hypothesis. The mean quantitative GRE score for students, whose family income was between $70,000 and $80,000, is significantly higher than then the average quantitative GRE score for the population.
Ho: µ Pop = µ sample
Ha: µ Pop < µ sample
Z = (mean1 - mean2)/standard error (SE) of difference between means
SEdiff = √(SEmean1^2 + SEmean2^2)
SEm = SD/√n
If only one SD is provided, you can use just that to determine SEdiff.
Find table in the back of your statistics text labeled something like "areas under normal distribution" to find the proportion/probability related to the Z score.
To perform a one-tailed hypothesis test, we will compare the sample mean to the population mean and determine the test statistic and the p-value.
First, let's calculate the test statistic using the formula:
Test Statistic = (Sample Mean - Population Mean) / (Standard Deviation / Square Root of Sample Size)
Given:
Population Mean (μ) = 500
Sample Mean (x̄) = 511
Standard Deviation (σ) = 100
Sample Size (n) = 400
Plugging in these values:
Test Statistic = (511 - 500) / (100 / √400)
= 11 / (100 / 20)
= 11 / 5
= 2.2
Therefore, the value of the test statistic is c. 2.2.
Next, we need to calculate the p-value associated with this test statistic. The p-value represents the probability of observing a test statistic as extreme as the one calculated, assuming the null hypothesis is true.
To calculate the p-value, we can use a statistical table or software. In this case, we are performing a one-tailed test, looking for scores significantly higher than the population mean. Since we don't have the exact p-value, we will approximate it using the table.
The p-value associated with the test statistic of 2.2 is approximately 0.0139.
Therefore, the p-value for the test statistic is c. 0.0139.
Finally, we compare the p-value to the significance level of 0.05 (or 5%). If the p-value is less than the significance level, we reject the null hypothesis; otherwise, we fail to reject the null hypothesis.
Since the p-value (0.0139) is less than the significance level (0.05), we can reject the null hypothesis.
Therefore, the answer is b. Yes, reject the null hypothesis. The mean quantitative GRE score for students whose family income was between $70,000 and $80,000 is significantly higher than the average quantitative GRE score for the population.