ST3004 Assessment Template

Date: February 7, 2026

Hypothesis testing

Part 1- Hypothesis testing in Research

References

Whitley, D., & Fuller-Thomson, E. (2017). African-American Solo Grandparents Raising Grandchildren: A Representative Profile of Their Health Status. Journal of Community Health, 42(2), 312–323. https://doi.org/10.1007/s10900-016-0257-8

All articles are available in full text in the Walden Library you may search by title or DOI. Be sure to provide an explanation for each of your answers and include supporting evidence from the text and/or Learning Resources.

Statistically non-significant relationship

Choose one characteristic listed on table 1 that does not show a statistical significance relationship with parenting status. Assume an alpha level of 0.05.

1. Write an appropriate null and alternative hypothesis test for this chi squared test

In Table 1, the selected characteristic that has no statistical significant relationship with parenting status is gender of child. The following are appropriate null and alternative hypotheses:

Null Hypothesis (H₀):

There is no association between the gender of the child and parenting status. Child gender distribution is the same for African-American solo grandparents and African-American single parents.

Alternative Hypothesis (H₁):

There is an association between the gender of the child and parenting status. The distribution of child gender differs between African-American solo grandparents and African-American single parents.

2. Explain why this relationship is NOT considered statistically significant. Explain how the p value provided supports your reasoning.

The relationship between gender of child and parenting status is not statistically significant because the p-value associated with the chi-square test is 0.47, which is greater than 0.05.

3. Write a conclusion statement to clearly explain decision for the results of this hypothesis test.

At 0.05 level of significance, we fail to reject the null hypothesis and conclude that there is no statistically significant association between the gender of the child and parenting status among African-American solo grandparents and African-American single parents.

4. Explain in the context of your hypothesis statements what it would mean to make a type II error.

A type II error occurs when the researcher fails to reject a false null hypothesis. Factors such as insufficient statistical power or limited sample size could lead to type II errors (Nestler & Salditt, 2024). In the context of the hypothesis test, a type II error would mean concluding that there is no association between child gender and parenting status when, in reality, there is a relationship between the variables.

Statistically significant relationship

Choose one characteristic listed on table 2 that does shows a statistical significance relationship with parenting status. Assume an alpha level of 0.05.

1. Write an appropriate null and alternative hypothesis test for this chi squared test

In Table 2, the selected characteristic that has no statistical significant relationship with parenting status is arthritis as a health indicator. The following are appropriate null and alternative hypotheses:

Null Hypothesis (H₀):

There is no association between arthritis status and parenting status.

Alternative Hypothesis (H₁):

There is an association between arthritis status and parenting status.

2. Explain why this relationship is considered statistically significant. Explain how the p value provided supports your reasoning.

The relationship is statistically significant because the p value is less than 0.05. There is enough statistical evidence to conclude that parenting status and arthritis status are related.

3. Write a conclusion statement to clearly explain decision for the results of this hypothesis test.

The null hypothesis was rejected because there is sufficient statistical evidence to conclude that there is a statistically significant relationship between parenting status and arthritis. This suggests that the prevalence of arthritis differs between the two parenting groups.

4. Explain in the context of your hypothesis statements what it would mean to make a type I error.

Type I error occurs when a true null hypothesis is rejected (Nestler & Salditt, 2024). In the current context, this would mean concluding that there is an association between arthritis and parenting status when, in reality, the relationship does not exist.

Part 2- Performing a hypothesis test

Comparing BMI by smoking status

Replace the questions below with your response to the following

This portion will make use of the body data set created in ST3001.

1. Explain which hypothesis test would be appropriate for this situation (assume population variances are not known and sample variances are not equal).

The appropriate hypothesis test for the situation is an independent two-sample t-test assuming unequal variances. The rationale for selecting the test is that the BMI is a continuous variable, and smoking status divides the participants into two independent groups (smokers and nonsmokers). The population variances are also unknown, and the sample variances are assumed to be unequal as stated in the instructions.

2. Explain if this is a left tailed, right tailed or two tailed test and justify your choice.

The research question is: Do smokers have a BMI that is greater than non-smokers?

Based on the research question, this is a right-tailed test because we are specifically testing if the mean BMI of the smokers is greater than that of nonsmokers. Therefore, the alternative hypothesis is directional.

3. Complete a hypothesis test in excel to test your claim. Copy and paste your excel output below.

Hypotheses:

Null hypothesis H₀ : The mean of smokers is less than or equal to that of non − smokers (μ_smokers ≤ μ_nonsmokers)

Alternative hypotheses: H₀ : The BMI of smokers is greater than that of non − smokers (μ_smokers > μ_nonsmokers)

Excel Output:

	BMI Smokers	BMI nonsmokers
Mean	28.10164	30.48525
Variance	30.91283	71.84761
Observations	61	61
Hypothesized Mean Difference	0
df	104
t Stat	-1.83648
P(T<=t) one-tail	0.034571
t Critical one-tail	1.659637
P(T<=t) two-tail	0.069143
t Critical two-tail	1.983038

4. Write a conclusion statement that makes use of excel data to justify which hypothesis is supported and why.

The results indicate that at the 0.05 significance level, t = -1.83648. Negative t-statistics indicate that the first group’s mean is smaller than the second group’s mean (West, 2021). Since the test statistic does not fall within the rejection region for a right-tailed test, and the p-value does not support the alternative hypothesis in the specified direction, we fail to reject the null hypothesis. The results indicate that the smokers do not have a statistically significant greater BMI than non-smokers. The negative t-value suggests that nonsmokers have a higher BMI than smokers.

5. State decision in context of the research question “Do smokers have a BMI that is greater than nonsmokers?”

There is insufficient statistical evidence to conclude that smokers have a BMI greater than nonsmokers.

Comparing weight by region of the country

Replace the questions below with your response to the following

This portion will make use of the body data set created in ST3001.

HINT: Weight data will need to be sorted by region like the process used to sort BMI by smoking status previously.

Comparing weight by region of the country

1. Explain which hypothesis test would be appropriate for this situation

The appropriate hypothesis test for the situation is a one-way ANOVA/ANOVA single-factor. The test is used to determine whether there are statistically-significant differences between the means of three or more independent groups (Gurvich & Naumova, 2021). In the question, weight is a continuous variable, and region is a categorical variable with four independent groups.

2. Complete a hypothesis test in excel to test your claim. Copy and paste your excel output below.

Null hypothesis H₀ : All four regions have equal average weights (μ₁ = μ₂ = μ₃ = μ₄)

Alternative hypotheses: H₁ : At least one region has a different mean weight

SUMMARY
Groups	Count	Sum	Average	Variance
Region 1	24	1890.4	78.76667	289.6136
Region 2	32	2570.8	80.3375	394.0185
Region 3	29	2483.2	85.62759	407.3335
Region 4	37	3109.7	84.04595	403.1781
ANOVA
Source of Variation	SS	df	MS	F	P-value	F crit
Between Groups	855.278	3	285.0927	0.75099	0.52389	2.681466
Within Groups	44795.44	118	379.6224
Total	45650.72	121

3. Write a conclusion statement that makes use of excel data to justify which hypothesis is supported and why.

At a 0.05 significant level, F(3,118) = 0.75099, p = .52389. Since the p-value is greater than the significance value, and the F statistic is not higher than the F critical value (0.75099 < 2.681466), we fail to reject the null hypothesis and conclude that there is no statistically significant difference in average weight among the four regions.

4. State decision in context of the research question “Do all four regions have equal average weights?”

There is insufficient statistical evidence to conclude that the four regions have different average weights. Therefore, the data support the conclusion that all four regions have equal average weights.

5. Create an Excel box plot of all 4 four regions side by side (like in STAT3001) and paste it here. Add a chart title and legend. Write a sentence or two on how this graph justifies your answer in question 4. Remember, these charts show the 5-number summary

The summary for the North region is: Min=48.0, Q₁=64.1, Median = 83.4, Q₃=91.1, Max=110.5. The summary for the South Region is: Min=48.5, Q₁=65.4, Median = 77.2, Q₃=93.9, Max=127.5. The summary for the East Region is: Min=52.1, Q₁=70.5, Median = 83.6, Q₃=95.6, Max=140.1. The summary for the West Region is: Min=46.3, Q₁=71.1, Median = 81.3, Q₃=116.0, Max=138.9.

Interpretation: The box plot illustrates the five-number summaries of weight across the four regions. The medians are similar across regions, with values of 83.4 in the North, 77.2 in the South, 83.6 in the East, and 81.3 in the West. The interquartile ranges overlap, with Q1 values ranging from 64.1 to 71.1 and Q3 values ranging from 91.1 to 116.0. This suggests that there is similar variability across regions. Although the East and West regions have higher maximum values with outliers, the overall distributions overlap considerably, and no region shows a clearly distinct weight distribution. This visual evidence supports the ANOVA results and justifies the conclusion that there is no statistically significant difference in average weight among the four regions.

References

Gurvich, V., & Naumova, M. (2021). Logical Contradictions in the One-Way ANOVA and Tukey–Kramer Multiple Comparisons Tests with More Than Two Groups of Observations. Symmetry, 13(8), 1387. https://doi.org/10.3390/sym13081387

Nestler, S., & Salditt, M. (2024). Comparing type 1 and type 2 error rates of different tests for heterogeneous treatment effects. Behavior Research Methods, 56(7), 6582–6597. https://doi.org/10.3758/s13428-024-02371-x

West, R. M. (2021). Best practice in statistics: Use the Welch t-test when testing the difference between two groups. Annals of Clinical Biochemistry International Journal of Laboratory Medicine, 58(4), 267–269. https://doi.org/10.1177/0004563221992088