You find out that the average 10th grade math score, for Section 6 of the local high school, is 87 for the 25 students in the class. The average test score for all 10th grade math students across the state is 85 for 1,800 students. The standard deviation for the state is 3.8.
Answer the following questions:
What z score do you calculate?
What is the area between the mean and the z score found in Appendix A of the textbook?
What does this mean about the probability of this test score difference occurring by chance? Is it less than 0.05?
To calculate the z-score, we can use the formula:
z = (x - μ) / σ
where x is the individual score, μ is the population mean, and σ is the population standard deviation.
In this case, the individual score is 87, the population mean is 85 (average test score for all 10th grade math students across the state), and the population standard deviation is 3.8.
Plugging in the values, we get:
z = (87 - 85) / 3.8
z = 2 / 3.8
z ≈ 0.526
So, the z-score is approximately 0.526.
To find the area between the mean and the z-score found in Appendix A of the textbook, we need to look up the z-score in the standard normal distribution table (also known as Appendix A in the textbook). The table gives you the area to the left of the z-score.
Assuming the value in Appendix A is z = 0.526, look it up in the table to find the corresponding area. Let's say the area is 0.7000 (this is just an example).
The area between the mean and the z-score would be:
1 - 0.7000
= 0.3000
This means that 30% of the scores fall between the mean and the z-score found in Appendix A.
Regarding the probability of this test score difference occurring by chance, if the area between the mean and the z-score is less than 0.05 (which it's not in this case), it would mean that there is a less than 5% chance of observing this test score difference by chance alone.
In this case, since the area is 0.3000 (greater than 0.05), it means that the probability of this test score difference occurring by chance is higher than 0.05 (or 5%).
To answer these questions, we first need to calculate the z score.
The z score measures how many standard deviations an individual score is from the mean. It allows us to compare scores from different distributions.
The formula to calculate the z score is: z = (X - μ) / σ
Where:
- X is the individual score,
- μ is the mean of the distribution,
- σ is the standard deviation of the distribution.
In this case, the individual score (X) is 87, the mean of the distribution (μ) is 85, and the standard deviation (σ) is 3.8.
Let's calculate the z score:
z = (87 - 85) / 3.8
z = 2 / 3.8
z ≈ 0.526
So, the calculated z score is approximately 0.526.
Now, to find the area between the mean and the z score in Appendix A of the textbook, we can use a standard normal distribution table, also known as the z-table.
The z-table provides the cumulative probability up to a given z score. We can use it to find the area between the mean and the z score.
Looking up the z score of 0.526 in the z-table, we find that the cumulative probability is approximately 0.6996.
This means that approximately 0.6996 or 69.96% of the scores in the distribution lie between the mean and the z score of 0.526.
Finally, to determine whether this test score difference occurring by chance is less than 0.05, we need to compare the obtained probability (0.6996) with the significance level of 0.05.
If the obtained probability is less than the significance level, then the test score difference can be considered statistically significant and unlikely to occur by chance. Otherwise, if the obtained probability is greater than the significance level, the difference may be due to chance.
In this case, the obtained probability of 0.6996 is greater than the significance level of 0.05. Therefore, we can conclude that the test score difference of 2 points is not statistically significant and could occur by chance.
Z = (score-mean)/SD
This should give you a good start:
Z = (mean1 - mean2)/standard error (SE) of difference between means
SEdiff = √(SEmean1^2 + SEmean2^2)
SEm = SD/√n