The Education Testing Service conducted a study to investigate differences between the scores of males and females on the SAT. The study identified a random sample of 562 females and 852 males that had achieved the same high score on the math portion of the test. That is, both females and males were viewed as having similar high abilities in Math. The SAT verbal scores for the two samples are:
Females Males
X1 = 547 X2 = 525
S1 = 83 S2 = 78
Do the data support the conclusion that give a population of females and a population of males with similar high mathematical abilities, the females will have a significantly higher verbal ability? Test at a 0.02 level of significance. What is your conclusion?
wow I wish you actually solved the problem out
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 your Z score.
To determine if the data supports the conclusion that females have a significantly higher verbal ability compared to males, we can conduct a two-sample t-test. The null hypothesis states that there is no difference in verbal ability between females and males, while the alternative hypothesis states that females have a significantly higher verbal ability.
Given the sample data, we have:
Sample size for females (n1) = 562
Sample size for males (n2) = 852
Sample mean for females (X1) = 547
Sample mean for males (X2) = 525
Sample standard deviation for females (S1) = 83
Sample standard deviation for males (S2) = 78
Based on these values, we can calculate the t-statistic and compare it to the critical t-value at a 0.02 level of significance.
The formula for calculating the t-statistic for independent samples is:
t = (X1 - X2) / sqrt((S1^2 / n1) + (S2^2 / n2))
Plugging in the values:
t = (547 - 525) / sqrt((83^2 / 562) + (78^2 / 852))
Calculating this expression gives us:
t ≈ (547 - 525) / sqrt((693289 / 562) + (6084 / 852))
≈ 22 / sqrt(1233 + 7.143)
Now, we need to find the critical t-value at a 0.02 level of significance with degrees of freedom (df) equal to the smaller sample size minus 1 (df = min(n1, n2) - 1).
Using a statistical table or software, the critical t-value for a two-tailed test with df = min(562, 852) - 1 ≈ 561 degrees of freedom and a significance level of 0.02 is approximately ±2.601.
Since the calculated t-value (approximately 22) is outside the range of the critical t-value (±2.601), we reject the null hypothesis.
Therefore, the data supports the conclusion that, given a population of females and a population of males with similar high mathematical abilities, the females have a significantly higher verbal ability.
To determine if the data supports the conclusion that females have a significantly higher verbal ability compared to males, we can conduct a hypothesis test using the information provided.
First, let's state the null and alternative hypotheses for the test:
Null Hypothesis (H0): There is no significant difference in verbal abilities between females and males with similar high mathematical abilities.
Alternative Hypothesis (HA): Females have a significantly higher verbal ability compared to males with similar high mathematical abilities.
Next, we need to determine the appropriate test statistic. Since we are comparing two independent samples and want to test the difference in means, we can use the two-sample t-test.
Now, let's calculate the test statistic value and the corresponding p-value using the provided data:
For females:
Sample mean (X1) = 547
Sample standard deviation (S1) = 83
Sample size (n1) = 562
For males:
Sample mean (X2) = 525
Sample standard deviation (S2) = 78
Sample size (n2) = 852
Now we can calculate the test statistic value using the formula:
t = (X1 - X2) / sqrt((S1^2/n1) + (S2^2/n2))
Plugging in the values, we get:
t = (547 - 525) / sqrt((83^2/562) + (78^2/852))
Now, we can find the p-value using the calculated test statistic and the degrees of freedom defined by (n1-1) + (n2-1) = (562-1) + (852-1) = 1412 degrees of freedom.
Using a t-table or a statistical software, we can find the p-value for the obtained test statistic value and the degrees of freedom. If the p-value obtained is less than the specified level of significance (0.02 in this case), then we reject the null hypothesis.
Based on the calculated test statistic and obtained p-value, we can make a conclusion. If the p-value is less than 0.02, we reject the null hypothesis and conclude that there is evidence to support the claim that females have significantly higher verbal ability compared to males with similar high mathematical abilities. If the p-value is greater than or equal to 0.02, we fail to reject the null hypothesis and conclude that there is not enough evidence to support the claim.
Please provide the t-value calculated using the formula, and I can guide you further in making the conclusion.