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Assessment 2: Hypothesis Testing, Correlation, Chi-Square, Dependent and Independent Samples T-Test.
Section A: Short Answer Questions
Question 1 In the chi-square test, when the expected value is less than 5, it implies that the assumption of the
test is violated. The chi-square assumes that the expected value for the frequencies is more than 5. An expected value of 5 compromised the validity and reliability of the results. When the expected value is less than 5, the following should be done;
1. A different statistical test should be used since the assumption of the chi-square test had been violated. The test will take into account the type of the variable and the objective of the test.
2. The sample size should be increased to obtain a higher expected value. This will be collecting more data for the study.
3. The adjusted cells with a low expected value should be combined to create larger groups. The categories will be collapsed to increase the expected count of cells.
4. Corrections such as Yates' continuity correction for 2*2 tables should increase the accuracy when using a small expected value.
Question 2 In this case, the most appropriate inferential statistical technique is correlation analysis. It examines the relationship between the Beck's Depression Inventory (BDI-II) scores and Drug Abuse Screening Test (DAST-10) scores among the inmates. The Beck's Depression Inventory (BDI-II) scores examine the level of depression among inmates, while the Drug Abuse Screening
Test (DAST-10) scores measure drug use. The two variables are continuous with an interval level
of measurement. Hence, the Pearson correlation coefficient will be used to assess the strength and direction of the relationship.
Question 3
The power of a test is the probability of rejecting the null hypothesis when it is false. A test with a higher power is regarded as very sensitive and better at detecting the true effects. The power of the test has an indirect relationship with Type II error. Type II error (β) arises when we fail to reject the null hypothesis when it is false. A higher power of test decreased the probability of making Type II errors. The power of the test is denoted by (1- β) while Type II error (β). Lowering the power of the test will increase the probability of committing Type II errors. Increasing the power of the test will decrease the probability of committing Type II errors.
The following actions will be implemented to have adequate power in a study.
1. Increasing the sample size. An increase in the sample size will increase the test's power by decreasing the probability of committing type II errors.
2. Increasing the effect size. An increase in the effect size will increase the test's power and reduce the likelihood of making Type II errors.
Question 4 There is a positive relationship between statistical significance, sample size and the effect size. It implies that a larger sample size will increase a test's statistical significance and power, making the test more likely to detect the true effect. Hence, a larger effect size increases the statistical significance, showing a strong relationship linking the variables. Also, increasing the sample size
makes it easier to detect smaller effect sizes. This suggests that a larger sample size allows for greater precision in estimating the true effect size.
Question 5 There is a difference between the chi-square test and the parametric tests. The two differ based on the hypotheses, data, and assumptions.
Hypotheses Chi-square tests are used to test hypotheses about the distribution of categorical data. On the other hand, parametric tests are used to test hypotheses about continuous data's means, differences or relationships.
Data The chi-square tests categorical variables that are either ordinal or nominal. Besides, the parametric tests assess continuous variables measured on an interval or ratio scale.
Assumptions For the chi-square tests, data is not normally distributed. It violates the assumption of normality. Besides, parametric tests assume normality of data and homogeneity of variance. Hence, it meets
the assumption of normality and equality of variance.
Section B: Calculations
Question 1 a. State the null and alternate hypotheses
The following null and alternative hypotheses will be tested.
Null hypothesis (H0): There is no relationship between employment status and anxiety level.
Alternative hypothesis (Ha): There is a relationship between employment status and anxiety level.
b. State the design requirements and assumptions needed to run this test
Design Requirements and Assumptions i) The variables should be categorical.
ii) The expected frequency for each cell should be at least 5.
iii) The sample size should be large.
iv) Data is drawn from a random sample.
c. State the decision rule
The decision rule for the chi-square test is the critical value. The critical value is compared with the test statistic; if the critical value is less than the test statistic, we reject the null hypothesis. We
fail to reject the null hypothesis if the critical value is greater than the test statistic.
d. Calculate the value of the test statistic and make a decision
The test statistic is computed using the following formula.
Employment Status
Anxiety Level
Low
Median
High
Total
Unemployed
13 (32.92)
40 (50.46)
67 (33.85)
120
Employed
68 (49.38)
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90 (75.69)
22 (50.77)
180
Retired
26 (24.69)
34 (37.85)
30 (25.38)
90
Total
107
164
110
390
(13-32.92)2/32.92 + (68-49.38)2/49.38 + (26-24.69)2/24.69 + (40-50.46)2/50.46+ (90-
75.69)2/75.69 +(34-37.85)2/37.85 + (67-33.85)2/33.85 + (22-50.77)2/50.77 + (30-25.38)2/25.38
= 12.05 + 7.02 +0.07+2.17+2.71+0.39+32.46+16.30+0.84
= 74.02
Degrees of freedom, df = (r-1)(c-1) = (3-1)(3-1) = 4
Using the chi-square table, we find the critical value, alpha =0.01/2 and df = 4.
Critical value = 14.8603
Since the critical value is less than the test statistic, , we reject the null hypothesis.
e. State the conclusion
Since the critical value is less than the test statistic, there is sufficient evidence to support the claim. Therefore, there is a significant relationship between employment status and anxiety level.
f. Calculate and interpret the effect size, if appropriate
We calculate Cramer's V, which is the effect size of the chi-square test.
V =
=
=
= 0.31
This is a moderate effect size. It implies that there is a moderate relationship between employment status and anxiety level.
Question 2
a. State the null and alternate hypotheses
Null hypothesis (H0): There is no significant difference in daily water usage before and after the campaign.
Alternative hypothesis (Ha): There is a significant difference in daily water usage before and after the campaign.
b. State the design requirements and assumptions needed to run this test
Design Requirements and Assumptions i) The data should be paired or matched
ii) The data should be normally distributed.
iii) The data should be randomly sampled from a population.
c. State the decision rule
The critical t-value will be compared with the test statistic (t-value). If the critical value is less than the test statistic, we reject the null hypotheses and decide there is a significant difference. We fail to reject the null hypothesis if the critical value is greater than the test statistic. Hence, there is no significant difference. The critical value will be tabulated at alpha =0.05/2 and df=n-
1= 7.
d. Calculate the value of the test statistic and make a decision
We will compute the paired t-test using the following formulae;
Where Is the difference in means, and µd is the population mean difference.
Household
Before Campaign
After campaign
Difference, d
1
298
294
4
2
254
251
3
3
224
226
-2
4
210
204
6
5
227
223
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4
6
262
255
7
7
241
229
12
8
280
265
15
= = = 6.125
= 6.125
The standard deviation for the difference, Sd, is calculated as;
Sd =
= 499
= =
= = 5.33
We plug the value in the equation to get;
= 7.50
= 7.5
Using the t-tables, we find the critical value with df=7 and alpha = 0.05.
Critical value = 2.3646
We reject the null hypothesis since the critical value (2.3646) is less than the test statistic (7.50).
e. State the conclusion
There is enough evidence to support the claim because the critical value is less than the test statistic. Therefore, we deduce a significant difference in daily water usage before and after the campaign.
f. Calculate and interpret effect size, if appropriate.
Effect size = = = 0.87
The effect size is large.
Question 3 a. State the null and alternate hypotheses
H0: There is no significant relationship between Social Media Addiction (SMA) and Jealousy.
Ha: There is a significant relationship between Social Media Addiction (SMA) and Jealousy.
b. State the design requirements and assumptions (assume assumptions have been met).
i)The two variables are variables that should be continuous.
ii)The data should be normally distributed.
ii) Homogeneity of variance.
iii) There is a linear relationship between Social Media Addiction (SMA) and Jealousy.
c. State the decision rule
We use the p-value to make decisions. If the p-value is less than 0.05, we reject the null hypothesis. Besides, when the p-value is greater than 0.05, we fail to reject the null hypothesis.
d. Compute the test statistic in Microsoft Excel and make a decision
SMA
MJS
SMA
1
MJS
0.205543
1
Tests statistic= =ABS(C2 / SQRT((1 - C2^2) / D2)) = 2.55
Test statistic, t = 2.55
p-value = =2 * T.DIST.2T(E2, D2)= 0.0238
= 0.0238
Since the p-value is less than 0.05, we reject the null hypothesis.
e. State the conclusion
There is enough evidence to support the claim because the p-value is less than 0.05. Therefore, we deduce that there is a significant relationship between Social Media Addiction (SMA) and Jealousy.
f. Calculate and interpret the effect size, if appropriate
The effect size is equal to the correlation coefficient, which is 0.21. It indicates that the effect size is small.
Question 4 a. State the null and alternate hypotheses
H0: There is no significant difference in the number of errors made on a test circuit between drivers who consumed two units of alcohol and those who did not
Ha: There is a significant difference in the number of errors made on a test circuit between drivers who consumed two units of alcohol and those who did not.
b. State the design requirements and assumptions (assume assumptions have been met)
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i)The dependent variable mis continuous
ii)The independent variable is two groups (experimental and control).
ii)The data is normally distributed
iii)There is homogeneity of variances.
iv)The data is drawn randomly from the population.
c. State the decision rule
In this case, we will be using the critical value rule. We compare the critical value and the test statistic. The null hypothesis is rejected because the critical value is less than the test statistic. The alpha = 0.05.
d. Calculate the value of the test statistic and make a decision
Manually
The test statistic is calculated using the following formulae;
Where the pooled variance,
The Group 1 (No Units Of Alcohol):
= 47/10 = 4.7
= 4.7
Variance, S21 = 3.12
The Group 2 (Two Units Of Alcohol):
= 74/10 = 7.4
= 7.4
Variance, S22 = 4.49
The pooled standard deviation,
,
= = 1.95
= 1.95
Hence, we have;
= = -3.10
= -3.10
We find the critical value using the t-table with 18 degrees of freedom and alpha = 0.05 since it's a two-tailed test.
Critical value = 2.1009
We reject the null hypothesis since the critical value (2.1009) is less than the test statistic (3.10).
Using Excel t-Test: Two-Sample Assuming Unequal Variances
Group 1
Group 2
Mean
4.7
7.4
Variance
3.122222
4.488889
Observations
10
10
Hypothesized Mean Difference
0
df
17
t Stat
-3.09485
P(T<=t) one-tail
0.003288
t Critical one-tail
1.739607
P(T<=t) two-tail
0.006577
t Critical two-tail
2.109816
We reject the null hypothesis since the critical value (2.109) is less than the test statistic (3.09).
e. State the conclusion
Since the null hypothesis is rejected, sufficient evidence supports the claim. Therefore, we conclude that there is a significant difference in the number of errors made on a test circuit between drivers who consumed two units of alcohol and those who did not.
f. Calculate and interpret effect size, if appropriate
The Cohen's d is computed using the following formula;
Cohen's d =
= = 0.71
= 0.71
The effect size is large.
Section C: Short Answer
a. Why the mean alone is not a useful statistic for significance testing
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The mean provided an average for all the values in the dataset. In statistical testing, the mean alone does not provide information regarding the variability or distribution of the data. Hence, it is not a useful statistic when used alone in significance testing. The test examines if there is a significant relationship or difference in the means of the variables. Also, it does not capture the uncertainty related to the estimate. Hence, it is not suitable for making statistical inferences.
b. The logic behind the central limit theorem The central limit theorem (CLT) states that the sample variable is normally distributed as the sample size is larger. The CLT is very important in hypothesis testing and providing inferences about the population means using a sample mean. It is applicable when the sample size is large enough or more than 30, and the observations are independent.
c. The logic of hypothesis testing
Hypothesis testing is a statistical procedure used to make inferences about a population based on a sample. It involved formulating the null and alternative hypotheses. The null hypothesis indicates no relationship or difference between the means, while the alternative hypothesis is the research claim. Data is gathered and analyzed using the appropriate statistical test. Then, the decision is made to reject or accept the null hypothesis. It indicates whether the results are statistically significant or not.
d. The rationale behind effect size statistics. The effect size shows the magnitude of the relationship or difference linking two or more variables. It provided vital insights into the practical significance. For example, when examining the relationship between Social Media Addiction (SMA) and Jealousy. We established that there is a positive relationship. The effect size was 0.31. This showed that there is a moderate relationship between Social Media Addiction (SMA) and Jealousy. Therefore, the effect size helps to understand the real-world implications of findings.
e.
The usefulness of non-parametric test statistics in social science research.
The non-parametric test is used in hypothesis testing when the assumptions of normality are violated. In social research, categorical variables are used to measure the participants' perceptions, opinions and demographic characteristics. Also, the non-parametric test is useful in social research when the sample size is small. The research may collect a small sample size when
there are limited resources. The non-parametric test is the most suitable test to be used. Therefore, the non-parametric test provides a flexible alternative for hypothesis testing and valuable insights if the assumptions of the parametric test assumption are not met.
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