1. On a standard measure of hearing ability, the mean is 300, and the standard deviation is 20. Provide the Z scores for persons whose raw scores are 340, 310, and 260. Provide the raw scores for persons whose Z scores on this test are 2.4, 1.5, and -4.5.
Z scores for raw scores 340, 310, and 260
Z score = (raw score - mean)/standard deviation
2 = (340 - 300)/20
0.5 = (310 - 300)/20
-2 = (260 - 300)/20
Raw scores for Z scores 2.4, 1.5, and -4.5
Raw score = mean +(standard deviation * Z score)
348 = 300 + (20 * 2.4)
330 = 300 + (20 * 1.5)
210 = 300 + (20 * -4.5)
2. Using the unit normal table, find the proportion under the standard normal curve that lies to the right of each of the following:
0.5 of the curve lies on the right hand(Heiman, 2011). Values on the right will be subtracted from 0.5 while values on the left side, that is values with a negative sign will be added to 0.5.
a. z = 1.00
0.5 - 0.3413 = 0.1587
b. z = -1.05
0.5 + 0.3531 = 0.8531
c. z = 0
0.5 + 0 = 0.5
d. z = 2.80
0.5 - 0.4974 = 0.0026
e. z = 1.96
0.5 - 0.0475 = 0.025
3. Suppose the scores of architects on a particular creativity test are normally distributed. Using a normal curve table (pp. 477-480 of the text), what percentage of architects have Z scores
For values above, we subtract the value from 1 and compute the percentage. Values below are computed to percentages without subtraction.
a. Above 0.10? 1 - 0.5398 = 0.4602 = 46.02%
b. Below 0.10? 0.5398 =53.98%
c. Above 0.20? 1 - 0.5793 = 0.4207 = 42.07%
d. Below 0.20? 0.5793 = 57.93%
e. Above 1.10? 1 - 0.8643 = 0.1357 = 13.57%
f. Below 1.10? 0.8643 = 86.43%
g. Above -0.10? 1 - 0.4602 = 0.5398 = 53.98%
h. Below -0.10? 0.4602 =46.02%
4. A statistics instructor wants to measure the effectiveness of his teaching skills in a class of 102 students (N = 102). He selects students by waiting at the door to the classroom prior to his lecture and pulling aside every third student to give him or her a questionnaire.
a. Is this sample design an example of random sampling?
Yes.
Explaination.
This a random sampling using the systematic sampling technique. Samples(students) are selected starting at a random point and continues at an interval (Thompson, 2013). The interval here is every third student. The starting point counts at the first student to attend the lesson. The first to be selected is the third one.
b. Assuming that all students attend his class that day, how many students will the instructor select to complete his questionnaire?
Divide the total number of students(102) by the sampling interval(3) to get 34 students. 102/3=34.
5. Suppose you were going to conduct a survey of visitors to your campus. You want the survey to be as representative as possible.
a. How would you select the people to survey?
Simple random sampling.
b. Why would that be your best method?
It an effective way of selecting samples from a larger population (Jackson, 2012) in this case many visitors. Each member has a chance of being chosen into the sample and it is by chance. The bias error is eliminated hence ending up with a sample representing all.
References
Heiman, G. W. (2011). Basic statistics for the behavioral sciences. Belmont, CA: Wadsworth Cengage Learning.
Jackson, S. L. (2012). Research methods and statistics: A critical thinking approach. Belmont, CA: Wadsworth Cengage Learning.
Thompson, S. K. (2013). Sampling. Hoboken, N.J: Wiley.