Hypothesis Testing: t-test & ANOVA

Task #1. Compare a mean score of one sample to a criterion 

The null and the alternative hypotheses were stated as follows.

Ho: The mean height of Navy recruits is 69.1 inches. 

Ha: The mean height of Navy recruits is higher than 69.1 inches 

If Ho is true and samples of size 64 are repeatedly drawn from the population of Navy recruits, sampling distribution of mean height will be a normal curve that centers on height 69.1 with a standard error of 0.

If Ha is true, the level of significance is ≤ 0.05. It is a one-tailed test in a positive direction. The critical z-score is 1.64.

From the analysis in tables 1 and 2, the mean difference is 1.583 (70.683 – 69.1), and the test statistic is 4.197, with the p-value being < .001 due to the hypothesis being directional.

Table 1. One-Sample T-Test
					t				df				p		Mean Difference		Cohen's d
Height					4.197				63				< .001		1.583		0.525
Note.  For the Student t-test, effect size is given by Cohen's d .
Note.  For the Student t-test, location parameter is given by mean difference d .
Note.  For the Student t-test, the alternative hypothesis specifies that the mean is different from 69.1.
Note.  Student's t-test.
Table 2. Descriptive s
		N		Mean		SD		SE
Height		64		70.683		3.017		0.377

The null hypothesis is rejected because the p-value is smaller than the stated level of significance. Additionally, t he absolute value of the test statistic 4.197 is larger than the absolute value of the critical score (1.64) , s o we reject Ho and accept Ha at the 95% confidence level. 

There is a reason to believe that the mean height for NAVY recruits is significantly higher than 69.1 inches. NAVY recruitments consider body physique such as height and weight, which resultantly affects the population's overall mean height.

  Task #2. Compare two independent groups (Gender) for consumer salary and spending 

The null and alternative hypotheses were stated as follows.

H1o: S alary between male and female are equal

H1a: S alary between male and female are different

H2o: S pending between male and female are equal

H2a: S pending between male and female are different

If Ho is true and samples of size 1000  are repeatedly drawn from the consumer population, the sampling distribution of the mean salary and spending will be a normal curve, which centers on 0 as an approximately normal t-distribution. We can assume equal variances when in the LEVENE’s test Si g. > 0.05. The number of degrees of freedom is df = 998 ( 400 + 600-2 ). The SE is for the difference in the means is calculated with the formula for the pooled variance. 

If H a is true , equal variances will be assumed. The level of significance is 0.05. The hypothesis is a two-tailed test, and the critical t-value is +/- 1.96.

The analysis was carried out in JASP, and the results were as shown in tables 3,4,5, and 6 and figure 1. The test effect showed that the mean difference was -15883.417 for salary and 52.175 for spending. The test statistic t was -11.400 for salary and 8.962 for spending. The p-values were < .001 for both salary and spending due to the hypothesis being 2-tailed. The effect size was given by Cohen’s d as -0.736 for salary and 0.578 for spending.

Independent Samples T-Test 

Table 3. Independent Samples T-Test
		Test		Statistic		df		p		Location Parameter		SE Difference		Effect Size
Salary		Student		-11.400		998.000		< .001		-15883.417		1393.276		-0.736
		Welch		-11.628		910.518		< .001		-15883.417		1365.996		-0.743
		Mann-Whitney		71014.000				< .001		-16700.000				-0.408
Spent		Student		8.962		998.000		< .001		467.592		52.175		0.578
		Welch		8.774		791.869		< .001		467.592		53.292		0.572
		Mann-Whitney		158731.000				< .001		480.000				0.323
Note.  For the Student t-test and Welch t-test, effect size is given by Cohen's d. For the Mann-Whitney test, effect size is given by the rank biserial correlation.
Note.  For the Student t-test and Welch t-test, location parameter is given by mean difference. For the Mann-Whitney test, location parameter is given by the Hodges-Lehmann estimate.

Assumption Checks 

Table 4. Test of Normality (Shapiro-Wilk)
				W		p
Salary		Female		0.965		< .001
		Male		0.991		< .001
Spent		Female		0.982		< .001
		Male		0.950		< .001
Note.  Significant results suggest a deviation from normality.

Table 5. Test of Equality of Variances (Levene's)
		F		df		p
Salary		7.738		1		0.006
Spent		6.375		1		0.012

Descriptives 

Table 6. Group Descriptives
		Group		N		Mean		SD		SE
Salary		Female		400		82969.250		20291.171		1014.559
		Male		600		98852.667		22404.684		914.667
Spent		Female		400		1797.125		859.135		42.957
		Male		600		1329.533		772.582		31.541

Descriptive Plots 

Figure 1. 

Salary 

Figure 2. 

Spent 

The decision is to reject the null hypotheses because the p-values for salary and spending are smaller than the stated level of significance. Additionally, the absolute value of the test statistic -11.4 for salary and 8.962 for spending is larger than the absolute value of the critical score of 1.96. We reject the null hypotheses with a 95% level of confidence.

There is a reason to believe that the male population's salary is higher than the female population. In contrast, the spending of the female population is higher than that of the male population. The difference in salaries is due to a gender pay gap attributed to discrimination in pay or the difference in education between men and women. Women spend more than men because of their lifestyle choices.

Task #3: Comparing more than two independent groups 

The null and alternative hypotheses were stated as follows.

H1o: There is no difference in golf driving distance between three temperature conditions (cool, mild, and warm) 

H1a: Golf driving distance in three weather conditions are not the same. 

H2o: There is no difference in golf driving distance among 5 different brands 

H2a: There are significant differences in golf driving distance among 5 different brands. 

If Ho is true and samples of size  300  are repeatedly drawn from the consumer population, sampling distribution of the mean distance will be a normal curve which centers on Yard = 0 with a standard error SE = 0 . 

If Ha is true, equal variances are assumed. The level of significance is 0.05. The hypothesis is a two-tailed test, and the critical t-value is 1.96.

The JASP analysis provided results that were shown in table 7,8,9,10,11,12,13 and 14 and figures 3 and 4. For the weather analysis, the column mean differences from post-hoc were -21.384 for cool and mild, -39.211 for cool and warm, and -17.827 for mild and warm. The t-statistics were 12.562, -23.034, and -10.472, as shown in table 10. The p values were < .001. For the comparison of brands, the t-values were shown in table 14, where the values varied from -14.248 to 8.342. The p-values were higher than 0.05, except for brand D and E.

ANOVA 

Table 7. ANOVA - Yards
Homogeneity Correction		Cases		Sum of Squares		df		Mean Square		F		p		ω²
None		Temp		77085.997		2.000		38542.998		266.021		< .001		0.639
		Residuals		43031.529		297.000		144.887
Brown-Forsythe		Temp		77085.997		2.000		38542.998		266.021		< .001		0.639
		Residuals		43031.529		292.497		147.118
Welch		Temp		77085.997		2.000		38542.998		282.340		< .001		0.639
		Residuals		43031.529		197.135		218.285
Note.  Type III Sum of Squares

Descriptives 

Table 8. Descriptives - Yards
Temp		Mean		SD		N
Cool		222.148		12.285		100
Mild		243.532		12.778		100
Warm		261.359		10.977		100

Assumption Checks 

Table 9. Test for Equality of Variances (Levene's)
F		df1		df2		p
1.670		2.000		297.000		0.190

Figure 3 

Q-Q Plot 

Post Hoc Tests 

Standard 

Table 10. Post Hoc Comparisons - Temp
				Mean Difference		SE		t		p tukey
Cool		Mild		-21.384		1.702		-12.562		< .001
		Warm		-39.211		1.702		-23.034		< .001
Mild		Warm		-17.827		1.702		-10.472		< .001
Note.  P-value adjusted for comparing a family of 3

Table 11. ANOVA - Yards
Homogeneity Correction		Cases		Sum of Squares		df		Mean Square		F		p		ω²
None		Brand		7702.436		4.000		1925.609		5.053		< .001		0.051
		Residuals		112415.090		295.000		381.068
Brown-Forsythe		Brand		7702.436		4.000		1925.609		5.053		< .001		0.051
		Residuals		112415.090		286.894		391.834
Welch		Brand		7702.436		4.000		1925.609		4.360		0.002		0.051
		Residuals		112415.090		147.250		763.430
Note.  Type III Sum of Squares

Descriptives 

Table 12. Descriptives - Yards
Brand		Mean		SD		N
A		237.902		19.222		60
B		242.515		17.365		60
C		244.583		18.149		60
D		236.242		20.691		60
E		250.490		21.836		60

Assumption Checks 

Table 13. Test for Equality of Variances (Levene's)
F		df1		df2		p
1.426		4.000		295.000		0.225

Q-Q Plot 

Post Hoc Tests 

Standard 

Table 14. Post Hoc Comparisons - Brand
				Mean Difference		SE		t		p tukey
A		B		-4.613		3.564		-1.294		0.695
		C		-6.682		3.564		-1.875		0.333
		D		1.660		3.564		0.466		0.990
		E		-12.588		3.564		-3.532		0.004
B		C		-2.068		3.564		-0.580		0.978
		D		6.273		3.564		1.760		0.399
		E		-7.975		3.564		-2.238		0.169
C		D		8.342		3.564		2.341		0.135
		E		-5.907		3.564		-1.657		0.462
D		E		-14.248		3.564		-3.998		< .001
Note.  P-value adjusted for comparing a family of 5

The decision from the temperature conditions is to reject the null hypotheses because the p-values were smaller than the stated level of significance. Additionally, the absolute values of the test statistics were larger than the absolute value of the critical score of 1.96. We reject the null hypothesis, with a 95% level of confidence.

The decision from the comparison of brands is to accept the null hypothesis because the p-values were higher than the stated level of significance. The values of the test statistic of the values were close to the critical score of 1.96. We accept the null hypothesis, with a 95% level of confidence. However, the brands D and E had p-values of less than the level of significance, and we reject the null hypothesis for these brands. The absolute values of the test statistics were larger than the absolute value of the critical score of 1.96. We reject the null hypothesis, with a 95% level of confidence for brands D and E.

Temperature conditions impact golf driving distance. The ball travels less distance during cold temperatures due to a higher density of air. The comparison of brands showed that it did not necessarily affect the driving distance as brands do not have any differences that could impact the driving distance. However, there were differences in brands D and E, which could be treated as an outlier.