Question

and with minitab please.

6. On Experiment Day when Dr. Lyth and Dr. Mallak conducted a group activity during class. Data was collected from both lecture sessions on each team's score and total table score for two different rounds. The data listed below show the total score in the second round for each lecture section. Section A-Table Score 200 140 180 160 150 200 140 200 140 Section B-Table Score 200 140 190 200 90 120 200 100 30 a) At 5% significance, test if "Section All performed better. Assume equal variance. b) In order to run the test in part (a), what is the assumption made about the data? Test the assumption. c) Given the conclusion from part (b). Is the test in part(a) valid?

Answer 1

	Section A	Section B
	200	200
	140	140
	180	190
	160	200
	150	90
	200	120
	140	200
	200	100
	140
	30
Mean	154	155
SD	50.59644	47.80914

(a and b)Assumptions and Test :

The provided sample means are shown below: X1 = 154 X2 = 155 Also, the provided sample standard deviations are: $1 = 50.5964 82 = 47.8091 and the sample sizes are ni = 10 and n2 = 8. (1) Null and Alternative Hypotheses The following null and alternative hypotheses need to be tested: Ho: 4 = 42 Ha: Mi > M2 This corresponds to a right-tailed test, for which a t-test for two population means, with two independent samples, with unknown population standard deviations will be used.

(2) Rejection Region Based on the information provided, the significance level is a = 0.05, and the degrees of freedom are df = 16. 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 right-tailed test is te = 1.746, for a = 0.05 and df = 16. The rejection region for this right-tailed test is R = {t:t > 1.746}. (3) Test Statistics Since it is assumed that the population variances are equal, the t-statistic is computed as follows: X1 - X2 (n-1) Tín, G+) 154 - 155 (10-1)50.59642+(8-1)47.8091 (1 10+8- 2 1 0 = = -0.043 + T & V

c)

(4) Decision about the null hypothesis Since it is observed that t = -0.043 <te = 1.746, it is then concluded that the null hypothesis is not rejected. Using the P-value approach: The p-value is p = 0.5168, and since p=0.5168 > 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 Mi is greater than um, at the 0.05 significance level.

Minitab > Stat > Basic Statistics > 2 sample t test

Enter Values as below

2-Sample t (Test and Confidence Interval) Samples in one column Samples: Subscripts: Samples in different columns First: Second: Summarized data Sample size: First: 10 Second: 8 Mean: 154 Standard deviation: 50.5964 155 47.8091 Assume equal variances Select Graphs... Options... Help | OK) Cancel

Select the options button and put values and hypothesis as below

2-Sample t (Test and Confidence Interval) Samples in one column Samples: Subscripts: 2-Sample t - Options Confidence level: 95.0 Test difference: 0.0 Alternative: greater than ndard riation: 5.5964 7.8091 Help OK Cancel POSTIC CUC VOTTOTIES Select Graphs... Options... Help OK Cancel

Clock ok to the options tab and next click ok to the main tab and then you will get an output as below

Minitab output :

- 11/10/2019 10:37:22 PM TWO-Sample T-Test and CI Sample 1 N 10 8 SE Mean StDev Mean 154.0 50.6 16 155.0 47.8 17 Difference = mu (1) - mu (2) Estimate for difference: -1.0 95% lower bound for difference: -41.9 T-Test of difference - 0 (vs>): T-Value = -0.04 Both use pooled StDev = 49.3963 P-Value = 0.517 DF = 16