PSYC 510
PSYC510 HOMEWORK DESCRIPTIVE GENERAL CONCEPT
/ ASERIES OF QUESTIONS ARE BASED ON THE
CONCEPT IN THIS WEAK ASSIGNNED READING AND
PRESENTATIONS.
HOMEWORK 6
z tests and single sample t-tests
When submitting this file, be sure the filename includes your full name, course and section.
Example: HW6_JohnDoe_510B01
Be sure you have reviewed this module/week’s lesson and presentations along with the practice
data analysis before proceeding to the homework exercises. Complete all analyses in SPSS, then
copy and paste your output and graphs into your homework document file. Answer any written
questions (such as the text-based questions or the APA Participants section) in the appropriate
place within the same file.
Every question is worth 2 points.
Part I: General Concepts
These questions are based on the concepts covered in this week’s assigned readings and
presentations.
Fill in the highlighted blanks with the best word or words.
1) You calculate a z-(score) or a single data point but you use a z-(test) if you are
comparing a sample mean with the population mean.
2) The critical value for a two-tailed one sample z- test is +/-: (1.96)
A researcher reports the results of a single-sample t-test as t(23) = 1.80. The degrees
3) of freedom for this t-test are (23), which means there were (24) total participants in
the researcher's sample.
4) We can most easily increase statistical power by: (increasing sample size)
As the sample size gets larger, the (distribution) begin to look more like the
5) (normal distribution).
Page 1 of 10
, PSYC 510
Given a population standard deviation of 25, calculate the standard deviation of the
mean (𝝈𝐗̅ ) given the following sample sizes (show all your work):
6) N = 15 Answer: 𝝈𝐗̅ = 25/(sqrt(15))= 6.455
7) N = 150 Answer: 𝝈𝐗̅ = 𝟐𝟓/(𝐬𝐪𝐫𝐭(𝟏𝟓𝟎))= 2.041
If you know the sample mean for worker satisfaction is 83.5, and the estimated standard
error of the mean is 9.5, and the sample size is 15, calculate the 95% confidence interval
for a non-directional test (show all your work):
8) Confidence Work: CI=Xbar +/- z(𝝈𝐗̅ ) = 83.5+/- 1.96(9.5)
Interval (lower and
Upper: 83.5+ (1.96*9.5)=102.12
upper limits):
Lower: 83.5- (1.96*9.5)= 64.88
Answer:
Upper: 102.12
Lower: 64.88
64.88-102.12
9) Written Answer: Based on our calculations, we are 95% confident that the
statement: population mean lies within 64.88 and 102.12.
Calculate the critical degrees of freedom and identify the critical t value for a single-
sample t test in each of the following situations, using p=.05 for all scenarios. Then, state
whether the null hypothesis would be accepted or rejected:
df= critical t = Accept or Reject Ho:
10) Two-tailed test, N = 25, t Answer: 24 Answer: 2.064 Answer: reject
= 1.85
df= critical t = Accept or Reject Ho:
11) One-tailed test, N = 25, Answer: 24 Answer: 1.711 Answer: accept
t = 1.85
A company decides to pilot a new program in one department (N=40) with the hope that
it may increase the number of sales per month. The mean number of sales per month for
those who participated in the new program is 51. The mean number of sales per month
for the overall population of sales personnel at this national company is 48 with a
standard deviation of 3.1.
Page 2 of 10
PSYC510 HOMEWORK DESCRIPTIVE GENERAL CONCEPT
/ ASERIES OF QUESTIONS ARE BASED ON THE
CONCEPT IN THIS WEAK ASSIGNNED READING AND
PRESENTATIONS.
HOMEWORK 6
z tests and single sample t-tests
When submitting this file, be sure the filename includes your full name, course and section.
Example: HW6_JohnDoe_510B01
Be sure you have reviewed this module/week’s lesson and presentations along with the practice
data analysis before proceeding to the homework exercises. Complete all analyses in SPSS, then
copy and paste your output and graphs into your homework document file. Answer any written
questions (such as the text-based questions or the APA Participants section) in the appropriate
place within the same file.
Every question is worth 2 points.
Part I: General Concepts
These questions are based on the concepts covered in this week’s assigned readings and
presentations.
Fill in the highlighted blanks with the best word or words.
1) You calculate a z-(score) or a single data point but you use a z-(test) if you are
comparing a sample mean with the population mean.
2) The critical value for a two-tailed one sample z- test is +/-: (1.96)
A researcher reports the results of a single-sample t-test as t(23) = 1.80. The degrees
3) of freedom for this t-test are (23), which means there were (24) total participants in
the researcher's sample.
4) We can most easily increase statistical power by: (increasing sample size)
As the sample size gets larger, the (distribution) begin to look more like the
5) (normal distribution).
Page 1 of 10
, PSYC 510
Given a population standard deviation of 25, calculate the standard deviation of the
mean (𝝈𝐗̅ ) given the following sample sizes (show all your work):
6) N = 15 Answer: 𝝈𝐗̅ = 25/(sqrt(15))= 6.455
7) N = 150 Answer: 𝝈𝐗̅ = 𝟐𝟓/(𝐬𝐪𝐫𝐭(𝟏𝟓𝟎))= 2.041
If you know the sample mean for worker satisfaction is 83.5, and the estimated standard
error of the mean is 9.5, and the sample size is 15, calculate the 95% confidence interval
for a non-directional test (show all your work):
8) Confidence Work: CI=Xbar +/- z(𝝈𝐗̅ ) = 83.5+/- 1.96(9.5)
Interval (lower and
Upper: 83.5+ (1.96*9.5)=102.12
upper limits):
Lower: 83.5- (1.96*9.5)= 64.88
Answer:
Upper: 102.12
Lower: 64.88
64.88-102.12
9) Written Answer: Based on our calculations, we are 95% confident that the
statement: population mean lies within 64.88 and 102.12.
Calculate the critical degrees of freedom and identify the critical t value for a single-
sample t test in each of the following situations, using p=.05 for all scenarios. Then, state
whether the null hypothesis would be accepted or rejected:
df= critical t = Accept or Reject Ho:
10) Two-tailed test, N = 25, t Answer: 24 Answer: 2.064 Answer: reject
= 1.85
df= critical t = Accept or Reject Ho:
11) One-tailed test, N = 25, Answer: 24 Answer: 1.711 Answer: accept
t = 1.85
A company decides to pilot a new program in one department (N=40) with the hope that
it may increase the number of sales per month. The mean number of sales per month for
those who participated in the new program is 51. The mean number of sales per month
for the overall population of sales personnel at this national company is 48 with a
standard deviation of 3.1.
Page 2 of 10
