Question

Q2. The data below represents the grades of students from 2 different classes.
Using appropriate measures, try determine which class performed better.
Make sure to explain your reasoning.
Class A	Class B
88	72
76	80
80	72
75	72
86	81
75	77
79	100
99	69
84	83
77	93
59	98
66	73
88	65
98	99
79	84
68	82
89	78
75	65
81	81
79	77

Answer 1

We can use t-test of independent means to check which class performed better:

first we calculate mean and stanadard deviation of both the samples:

The sample size is n=20. The provided sample data along with the data required to compute the sample mean and sample variance s² are shown in the table below:

	class A	class A2
	88	7744
	76	5776
	80	6400
	75	5625
	86	7396
	75	5625
	79	6241
	99	9801
	84	7056
	77	5929
	59	3481
	66	4356
	88	7744
	98	9604
	79	6241
	68	4624
	89	7921
	75	5625
	81	6561
	79	6241
Sum =	1601	129991

The sample mean is computed as follows:

Also, the sample variance s² is:

Therefore, the sample sandartd deviation ss is

The sample size is n = 20 The provided sample data along with the data required to compute the sample mean and sample variance s² are shown in the table below:

	class B	class B2
	72	5184
	80	6400
	72	5184
	72	5184
	81	6561
	77	5929
	100	10000
	69	4761
	83	6889
	93	8649
	98	9604
	73	5329
	65	4225
	99	9801
	84	7056
	82	6724
	78	6084
	65	4225
	81	6561
	77	5929
Sum =	1601	130279

The sample mean is computed as follows:

Also, the sample variance s² is

Therefore, the sample sandartd deviation ss is

Now, we apply t-test of independent samples:

(1) Null and Alternative Hypotheses

The following null and alternative hypotheses need to be tested:

Ho: μ1​ = μ2​ or there is no significant difference in the perfomance of both the classes.

Ha: μ1​ ≠ μ2​ or there is a significant difference in the perfomance of both the classes.

(2) Rejection Region

Based on the information provided, the significance level is α=0.05, and the degrees of freedom are df=38. In fact, the degrees of freedom are computed as follows, assuming that the population variances are equal:

Hence, it is found that the critical value for this two-tailed test is tc​=2.024, for α=0.05 and df=38.

(3) Test Statistics

The provided sample means are shown below:

Also, the provided sample standard deviations are:

and the sample sizes are and

Since it is assumed that the population variances are equal, the t-statistic is computed as follows:

t=

t=0

(4) Decision about the null hypothesis

Since it is observed that ∣t∣=0≤tc​=2.024, it is then concluded that the null hypothesis is not rejected.

Using the P-value approach: The p-value is p=1, and sincep=1≥0.05, it is concluded that the null hypothesis is not rejected.

(5) Conclusion

It is concluded that the null hypothesis Ho is not rejected. Therefore, there is not enough evidence to claim that the population mean μ1​ is different than μ2​, at the 0.05 significance level.

Since we accept the null hypothesis hence,there is no significant difference in the perfomance of both the classes,both the classes perfomed equally better.

please rate my answer and comment for doubts.