HOMEWORK: Z TESTS AND SINGLE SAMPLE T – TESTS ASSIGNMENT
INSTRUCTIONS
OVERVIEW
This Homework: Z-Tests & Single Sample T-Tests Assignment is designed to assess your understanding of the concepts and applications covered thus far in this course. This module introduces two inferential tests – the z test and single sample t test; both of which are used to determine whether a sample differs from a population. These tests are commonly used in psychology, education, business, and other fields. This Homework: Z-Tests & Single Sample T-Tests Assignment assesses your knowledge, skills, applications, and presentation of these tests using professional conventions.
INSTRUCTIONS
Be sure you have reviewed this module’s Learn section before completing this Homework: Z- Tests & Single Sample T-Tests Assignment. This Homework: Z-Tests & Single Sample T- Tests Assignment is worth 60 points. Each question is worth 3 points. Six points are awarded for mechanics/structure. · Part I contains general concepts from this module’s Learn section. · Part II requires use of SPSS. You will have to take screen shots and/or copy and paste from your SPSS to place answers within this file. Make sure you only insert relevant and legible images. · Part III is the cumulative section. These may include short answer and/or use of SPSS but will review material from previous module(s). · Directions for each subsection are provided in the top of each table (in the blue shaded areas). · Answers should be placed where indicated (wherever there is “ ANSWER ”). · Submit the file as a WORD document (.doc or .docx). Make sure the filename of your submission includes your full name, course and section. o Example: HW6_JohnDoe_510B Make sure to check the Homework Grading Rubric before beginning this Homework: Z-Tests & Single Sample T-Tests Assignment.
Part I: General Concepts
These questions are based on the concepts covered in this module’s assigned readings and presentations.
Answer the following questions using your own words.
- What is the critical value for a one-tailed z test? For a two-tailed z test? Which is more likely to result in a rejection of the null hypothesis? Why?
A one-tailed z test has a critical value of +/-1. A two-tailed z test’s critical value is +/-1.
As long as the effect is in the indicated direction, it is easier to reject the null hypothesis with a one-tailed test than with a two-tailed test. One-tailed tests, on the other hand, have lower Type II error rates and more power than two-tailed tests. 2. Why does tcv****. Change when the sample size changes? What must be computed to determine tcv .? Effect of Sample Size: Increasing the sample size, for example, improves the normality of the t- distribution. Because tcv moves closer to 0 for a normal distribution, the mean difference of the sample is less likely to be farther / distant from the 0. In the case of a smaller sample size, the opposite is true.
The degree of freedom and the alpha or significance value are two requirements for computing tcv 3. Henry performed a two-tailed test for an experiment in which N = 24. He could not find his table of t critical values, but he remembered the tcv at df = 13. He decided to compare his tobt with this tcv****. Is he more likely to make a Type I or a type II error in this situation? Briefly discuss why. The small sample size has more chances of Type II error.
We can clearly see that at df = 13 sample size = 14 which is lesser than the sample size = 24 with df =
- tcv at df = 13 is larger than the tcv value of df = 23. Therefore, Henry is more likely to make Type II
error because he is likely to accept the null hypothesis instead of rejecting it.
● Tcv(23) = ±2. ● Tcv(13) = ±2.
Given a population standard deviation of 12, answer the following two questions. 4) Calculate the standard error of the mean ( given the following two sample sizes (show all your work and round to two decimal places): N 1 = 20 N 2 = 70
WORK:
N 1 =
SE = SD / Square Root (SR) of N SE = 12 / SR (20) SE = 12 / 4.
N 2 =
SE = SD / SR(N2)
SE = 12 / SR(70)
SE = 12 / 8.
ANSWER
N 1 = 2.
N 2 = 1.
5) How does the sample size change the standard error of the mean? What does this mean for statistical power?
The standard error measures the distribution’s scattering. As the sample size grows larger, the scattering or dispersion reduces, and vice versa. Because the distribution’s mean is near to the population’s mean, sample size is inversely associated with the
(SD) / Square Root (SR) of N
SE = 6 / SR (75) SE = 6 / 8 = 0. SE = 0. 11) Conduct the z-test. Show your work (use at least two decimal places throughout). What is the zobt score?
ANSWER
Zobt =Sample Mean – Population Mean / SD √N Zobt = 87 – 85 / 6 √ Zobt = 2 / 0. Zobt = 2. 12) Write your conclusion using complete sentences in APA format. Don’t forget to include the statistical statement and whether you reject or fail to reject the null hypothesis.
ANSWER
Z-scores were computed for health scores of the employees. The health score of exercising-employees (M = 87) was significantly greater than the all-employees (M = 85), zobt = 2 > zcv = 1, p = .004. Hence null hypothesis was rejected.
Part II: SPSS Application
These questions require the use of SPSS. Remember you must submit all of your work within this word document. You will need to take a screen shot of your data view if necessary, or copy and paste your output into the spaces below. Remember to report the exact p value provided by SPSS output – simply reporting p<.05 or p>.05 is not acceptable (unless SPSS output states p= – in that case you can report p<.001).
Exercise 1
Spatial ability was quantified using a sample of the general population (N=15). Their scores are shown below, where higher scores represent better spatial ability. Enter this data into SPSS to answer the following questions.
52 59 63 65 58
55 62 63 53 59
57 61 60 59 57
13)
Run descriptive statistics for general population data provided above (make sure to include the mean and confidence interval by using the “Explore” option as shown in this module’s presentation).
ANSWER SPSS output of descriptive statistics Descriptives Statisti c
Std. Error Spatial_Abilit y
Mean 58.
95% Confidence Interval for Mean
Lower Bound
56.
Upper Bound
60.
5% Trimmed Mean 58. Median 59. Variance 13. Std. Deviation 3. Minimum 52 Maximum 65 Range 13 Interquartile Range 5 Skewness -. Kurtosis – 1.
14) You want to compare your sample of the general population to individuals who
listen to classical music, but you only know their mean ( m = 60). Run a single sample t test. Paste the output below.
Grit GPA 6 4. 4 3. 3 2. 5 3. 4 3. 5 3. 3 2. 1 2. 3 3. 2 2. 5 3. 3 2. 5 2. 6 2.
16) Paste appropriate SPSS output.
ANSWER Model Summaryb
Model R
R
Square
Adjusted R Square
Std. Error of the Estimate 1 .523a .273 .213 1. a. Predictors: (Constant), GPA b. Dependent Variable: Grit
ANOVAa
Model
Sum of Squares df
Mean Square F Sig. 1 Regression 8 1 8 4 .055b Residual 21 12 1. Total 29 13 a. Dependent Variable: Grit b. Predictors: (Constant), GPA
Coefficientsa
Model
Unstandardized Coefficients
Standardized Coefficients Std. Error Beta t Sig 1 (Constant )
– 2 -.
GPA 1 .733 .523 2.
a. Dependent Variable: Grit
17) Paste appropriate SPSS graph. ANSWER
18) Write a Results section in current APA style describing the outcome using complete sentences. Make sure you report the statistical notation in APA format and inclue all relevant information.
ANSWER
A simple linear regression was computed to predict the Grit based on GPA. A non-significant
regression equation was found F(1, 12) = 4, p = .055, R 2 = .273, so GPA explained 27%
variance in Grit. Predicted Grit was equal to -0 + 1(GPA), meaning that average Grit
increased by 1 units with each increase in GPA unit, a scatter plot was also drawn to depict
the same. Hence it was concluded that GPA failed to predict the Grit at 5% level of
significance.
Submit this assignment by 11:59 p. (ET) on Sunday of Module 6. Remember to name the file appropriately.
Top of Form
Bottom of Form
A one-tailed z test has a critical value of +/-1.645
A two-tailed z test’s critical value is +/-1.96
As long as the effect is in the indicated direction, it is easier to reject the null hypothesis with a one-tailed
test than with a two-tailed test. One-tailed tests, on the other hand, have lower Type II error rates and
more power than two-tailed tests.
2. Why does tcv. Change when the sample size changes? What must be computed to determine tcv.?
Effect of Sample Size: Increasing the sample size, for example, improves the normality of the t-
distribution. Because tcv moves closer to 0 for a normal distribution, the mean difference of the sample is
less likely to be farther / distant from the 0. In the case of a smaller sample size, the opposite is true.
The degree of freedom and the alpha or significance value are two requirements for computing tcv
3. Henry performed a two-tailed test for an experiment in which N = 24. He could not find his
table of t critical values, but he remembered the tcv at df = 13. He decided to compare his tobt
with this tcv. Is he more likely to make a Type I or a type II error in this situation? Briefly
discuss why.
The small sample size has more chances of Type II error.
We can clearly see that at df = 13 sample size = 14 which is lesser than the sample size = 24 with df =
23. tcv at df = 13 is larger than the tcv value of df = 23. Therefore, Henry is more likely to make Type II
error because he is likely to accept the null hypothesis instead of rejecting it.
● Tcv(23) = ±2.0687
● Tcv(13) = ±2.1604
Given a population standard deviation of 12, answer the following two questions.
4) Calculate the standard error of
the mean ( given the following two
sample sizes (show all your work
and round to two decimal places):
N1 = 20
N2 = 70
WORK:
N1 =
SE = SD / Square
Root (SR) of N1
SE = 12 / SR (20)
SE = 12 / 4.472
N2 =
SE = SD / SR(N2)
SE = 12 / SR(70)
SE = 12 / 8.367
ANSWER
N1 = 2.68
N2 = 1.43
5) How does the sample size change
the standard error of the mean?
What does this mean for statistical
power?
The standard error measures the distribution’s scattering. As the
sample size grows larger, the scattering or dispersion reduces,
and vice versa. Because the distribution’s mean is near to the
population’s mean, sample size is inversely associated with the
sample’s standard error.
The mean difference in a comparison, i.e., population mean and
sample mean, is at a minimum, indicating that the subjects
function as their own controls and do not enable variations
between subjects to enter the study. The statistical power of the
study grows when the differences are reduced.
Calculate the critical degrees of freedom and identify the critical values for the following tests,
using p = .05 and the tables in your e-book for all scenarios (do NOT round). Then, state whether
the null hypothesis would be rejected or failed to be rejected.
6) Single sample t-test:
two-tailed, N = 20, t =
2.05
df=
ANSWER
df= N-1=20-
1=19
critical t =
ANSWER
critical t = 2.093
Reject or Fail to Reject Ho: failed to
reject.
ANSWER
t < tcv = 2.05 < 2.093, hence H0 failed
to be rejected.
7) Single sample t-test:
one-tailed, N = 20, t =
2.05
df= n – 1
df = 20 – 1
ANSWER
df = 19
critical t = 1.729
ANSWER
tcv = 1.729
Reject or Fail to Reject Ho: reject
ANSWER
t > tcv = 2.05 > 1.729, hence H0
rejected.
8) Pearson’s r
correlation: two-tailed,
N = 15, r = 0.47
df = n – 2
df = 15 – 2
ANSWER
df = 13
critical t = 2.1604
ANSWER
tcv = 2.1604
Reject or Fail to Reject Ho: reject.
ANSWER
t > tcv = 2.1751 > 2.1604, hence H0
rejected.
9) Pearson’s r
correlation: one-tailed,
N = 15, r = 0.47
df= n – 2
df = 15 – 2
ANSWER
df = 13
critical t = 1.7709
ANSWER
tcv = 1.7709
Reject or Fail to Reject Ho: reject.
ANSWER
t > tcv = 2.1751 > 1.7709, hence H0
rejected.
A company decides to offer a monetary incentive for employees who log a specified
number of hours spent exercising in an attempt to improve employee health. The average
health score (higher = better) of the 75 employees who logged enough hours was 87. The
mean health score for all employees (population) was 85, with a population standard
deviation of 6.5.
10) Calculate the standard error of the mean. ANSWER
Standard Error (SE) = Standard Deviation