Question

I am doing a T test for this problem and trying to see how to do it and which t test is best. Here is the data

Infractions of two distict groups of freshman and sophmore

fresman

7

4

8

7

4

8

9

7

Sophmore

5

5

4

5

6

4

5

3

Answer 1

Solution:-

Freshman		Sophmore
Mean	6.75	Mean	4.625
Standard Error	0.647798	Standard Error	0.323899
Median	7	Median	5
Mode	7	Mode	5
Standard Deviation	1.832251	Standard Deviation	0.916125
Sample Variance	3.357143	Sample Variance	0.839286
Kurtosis	-0.49181	Kurtosis	0.420824
Skewness	-0.81286	Skewness	-0.48772
Range	5	Range	3
Minimum	4	Minimum	3
Maximum	9	Maximum	6
Sum	54	Sum	37
Count	8	Count	8

State the hypotheses. The first step is to state the null hypothesis and an alternative hypothesis.

Null hypothesis: u₁ = u ₂
Alternative hypothesis: u₁$\neq$ u₂

Note that these hypotheses constitute a two-tailed test. The null hypothesis will be rejected if the difference between sample means is too big or if it is too small.

Formulate an analysis plan. For this analysis, the significance level is 0.05. Using sample data, we will conduct a two-sample t-test of the null hypothesis.

Analyze sample data. Using sample data, we compute the standard error (SE), degrees of freedom (DF), and the t statistic test statistic (t).

SE = sqrt[(s₁²/n₁) + (s₂²/n₂)]
SE = 0.7243
DF = 14
t = [ (x₁ - x₂) - d ] / SE

t = 2.934

where s₁ is the standard deviation of sample 1, s₂ is the standard deviation of sample 2, n₁ is the size of sample 1, n₂ is the size of sample 2, x₁ is the mean of sample 1, x₂ is the mean of sample 2, d is the hypothesized difference between the population means, and SE is the standard error.

Since we have a two-tailed test, the P-value is the probability that a t statistic having 14 degrees of freedom is more extreme than -2.934; that is, less than -2.934 or greater than 2.934.

Thus, the P-value = 0.011

Interpret results. Since the P-value (0.011) is less than the significance level (0.05), we have to reject null hypothesis.