The question:
If α =.05, 2-tail, would you consider responses of 9 or more words to be a rare occurrence? Why or why not? How about a score of 10? The mean is 6 and the standard deviation is 2.
My work:
Drew a distribution with .025 and .025 at both ends and .95 in between.
Then:
z9=9-6/2=3/2=1.5 (Located under z-score table as .4332)
z10=10-6/2=4/2=2.0 (Located under z-score table as .4772).
Then:
.4772+.4332=.9104=.91
Then: .95-.91=.04
Did I use the correct mathematical approach?
Yes, you used the correct mathematical approach to calculate the z-scores for the given values. The formula for calculating the z-score is (x - μ) / σ, where x is the value in question, μ is the mean, and σ is the standard deviation.
For the first part of the question, you calculated the z-score for a response of 9 words. You substituted the values into the formula as follows:
z9 = (9 - 6) / 2 = 3 / 2 = 1.5
You then looked up the corresponding area in the z-score table and found it to be approximately 0.4332.
Similarly, for the second part of the question, you calculated the z-score for a response of 10 words:
z10 = (10 - 6) / 2 = 4 / 2 = 2.0
You looked up the corresponding area in the z-score table and found it to be approximately 0.4772.
After finding these values, you correctly added them together:
0.4772 + 0.4332 = 0.9104
This represents the total area under the curve beyond a z-score of 2 (which corresponds to a score of 10 or higher).
To determine if these scores are considered rare occurrences, you compared the result to the significance level (α) of 0.05. Since 0.9104 is greater than 0.05, you would not consider responses of 9 or more words to be a rare occurrence.
Additionally, you calculated the area under the curve up to a z-score of 1.5:
0.9500 - 0.9104 = 0.0396
This represents the percentage of responses that fall between the mean and a z-score of 1.5, which is approximately 3.96%. Therefore, a score of 10 would be considered a rare occurrence, as it falls within the top 3.96% of the distribution.