The university conducted a study to observe the efficiency of students to see whether students do better when they study all at once or in intervals. One group of 12 participants participated in the survey after studying for one hour continuously. The other group of 12 participated in the survey after studying for three twenty-minute sessions. As per the survey, results group 1 had a mean score of 75 and a variance of 120. Second group had a mean score of 86 and a variance of 100. Assuming the normal populations, independent samples, and equal population variances conditions hold, are the mean test scores of these two groups significantly different at the 0.05 level?

Question
Comparing two groups
asked 2020-11-26
The university conducted a study to observe the efficiency of students to see whether students do better when they study all at once or in intervals.
One group of 12 participants participated in the survey after studying for one hour continuously.
The other group of 12 participated in the survey after studying for three twenty-minute sessions.
As per the survey, results group 1 had a mean score of 75 and a variance of 120.
Second group had a mean score of 86 and a variance of 100. Assuming the normal populations, independent samples, and equal population variances conditions hold, are the mean test scores of these two groups significantly different at the 0.05 level?

Answers (1)

2020-11-27
The objective is to check whether the mean test scores of the two groups significantly differ at the 0.05 level.
Let \(\mu_{1}\) denote the mean of the participants participated in the survey after studying for one hour continuously.
Let \(\mu_{2}\) denote the mean of the participants participated in the survey after studying for three twenty-minute sessions.
The null hypothesis is that there is no difference between the mean test of scores between the two groups. That is,
\(H_{0}\ :\ \mu_{1} = mu_{2}\)
The alternative hypothesis is that there is a difference between the mean test of the score between the two groups. That is,
\(H_{1}\ :\ \mu_{1}\ \neq\ \mu_{2}\)
The level of significance is,
\(\alpha = 0.05\)
Group 1:
Sample mean, \(\bar{x}_{1}= 75\)
Sample variance, \(S^{\frac{2}{1}} = 120\)
Sample Size, \(n1 = 12\)
Group 2:
Sample mean, \(\bar{x}_{2} = 86\)
Sample variance, \(S^{\frac{2}{2}} = 100\)
Sample Size, \(n2 = 12\)
\(t=\frac{(\bar{x}_{1}\ -\ \bar{x}_{2})\ -\ (\mu_{1}\ -\ \mu_{2})}{\sqrt{S^{\frac{2}{p}}\left(\frac{1}{n}\ 1\ +\ \frac{1}{n}\ 2\right)}}\)
The formula for pooled variance is,
\(S^{\frac{2}{p}}=\frac{(n1\ -\ 1)S^{\frac{2}{1}}\ +\ (n2\ -\ 1)S^{\frac{2}{2}}}{n1\ +\ n2\ -\ 2}\)
\(=\frac{(12\ -\ 1)120\ +\ (12\ -\ 1)100}{12\ +\ 12\ -\ 2}\)
\(=\frac{11(120)\ +\ 11(100)}{24\ -\ 2}\)
\(=\frac{1320\ +\ 1100}{22}\)
\(= \frac{2420}{22}\)
\(= 110\)
The test statistic is computed below.
\(t=\frac{(\bar{x}_{1}\ -\ \bar{x}_{2})\ -\ (\mu_{1}\ -\ \mu_{2})}{\sqrt{S^{\frac{2}{p}}\left(\frac{1}{n}\ 1\ +\ \frac{1}{n}\ 2\right)}}\)
\(=\frac{75\ -\ 86\ -\ 0}{\sqrt{110\left(\frac{1}{12}\ +\ \frac{1}{12}\right)}}\)
\(=\frac{-11}{\sqrt{110\left(0.08333\ +\ 0.08333\right)}}\)
\(=\frac{-11}{\sqrt{110\left(166667\right)}}\)
\(=\frac{-11}{\sqrt{18.33333}}\)
\(=\frac{-11}{4.281744}\)
\(=\ -2.569047\)
The degrees of freedom is,
\(df = n1\ +\ n2\ -\ 2 = 12\ +\ 12\ -\ 2 = 24\ -\ 2 = 22\)
Computation of p value:
from the alternative hypothesis,it is clear that the test is a two- tailed test.
t represents the test statistic and df is the degrees of freedom p value
\(= t\ dist (t,\ df\ tails)\)
\(= t\ dist(2.569047,\ 22,\ 2)\)
\(- 0.017503\)
Decision rule:
Reject the null hypothesis if p value is less than the level of significance.
That is, p value \(<\ 0.05\)</span>
Conclusion:
Here the p value is 0.017503 is less than 0.05. So, reject the null hypothesis and conclude that there is a significant difference between the means of the two groups.
0

Relevant Questions

asked 2020-10-23
The table below shows the number of people for three different race groups who were shot by police that were either armed or unarmed. These values are very close to the exact numbers. They have been changed slightly for each student to get a unique problem.
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Black - 543
White - 1176
Hispanic - 378
Total - 2097
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White - 67
Hispanic - 38
Total - 165
Total:
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This just means that the percentage of times that both things happen equals the individual percentages multiplied together (Only if they are Independent of each other).
Therefore, if a person's race is independent of whether they were killed being unarmed then the percentage of black people that are killed while being unarmed should equal the percentage of blacks times the percentage of Unarmed. Let's check this. Multiply your answer to part a (percentage of blacks) by your answer to part b (percentage of unarmed).
Remember, the previous answer is only correct if the variables are Independent.
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If answer c is "significantly different" than answer d, then that means that there could be a different percentage of unarmed people being shot based on race. We will check this out later in the course.
Let's compare the percentage of unarmed shot for each race.
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If you compare answers d, e and f it shows the highest percentage of unarmed people being shot is most likely white.
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This is because there are more white people in the United States than any other race and therefore there are likely to be more white people in the table. Since there are more white people in the table, there most likely would be more white and unarmed people shot by police than any other race. This pulls the percentage of white and unarmed up. In addition, there most likely would be more white and armed shot by police. All the percentages for white people would be higher, because there are more white people. For example, the table contains very few Hispanic people, and the percentage of people in the table that were Hispanic and unarmed is the lowest percentage.
Think of it this way. If you went to a college that was 90% female and 10% male, then females would most likely have the highest percentage of A grades. They would also most likely have the highest percentage of B, C, D and F grades
The correct way to compare is "conditional probability". Conditional probability is getting the probability of something happening, given we are dealing with just the people in a particular group.
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You can see by the answers to part g and h, that the percentage of blacks that were unarmed and killed by police is approximately twice that of whites that were unarmed and killed by police.
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State the null and alternate hypotheses.
\(H0:?_{1}^{2}=?_{2}^{2},H1:?_{1}^{2}>?_{2}^{2}H0:?_{1}^{2}=?_{2}^{2},H1:?_{1}^{2}\neq?_{2}^{2}H0:?_{1}^{2}=?_{2}^{2},H1:?_{1}^{2}?_{2}^{2},H1:?_{1}^{2}=?_{2}^{2}\)
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What are the degrees of freedom?
\(df_{N} = ?\)
\(df_{D} = ?\)
What assumptions are you making about the original distribution?
The populations follow independent normal distributions. We have random samples from each population.The populations follow dependent normal distributions. We have random samples from each population.The populations follow independent normal distributions.The populations follow independent chi-square distributions. We have random samples from each population.
(c) Find or estimate the P-value of the sample test statistic. (Round your answer to four decimal places.)
(d) Based on your answers in parts (a) to (c), will you reject or fail to reject the null hypothesis?
At the ? = 0.05 level, we fail to reject the null hypothesis and conclude the data are not statistically significant.At the ? = 0.05 level, we fail to reject the null hypothesis and conclude the data are statistically significant. At the ? = 0.05 level, we reject the null hypothesis and conclude the data are not statistically significant.At the ? = 0.05 level, we reject the null hypothesis and conclude the data are statistically significant.
(e) Interpret your conclusion in the context of the application.
Reject the null hypothesis, there is sufficient evidence that the population variance is larger in the old thermostat temperature readings.Fail to reject the null hypothesis, there is sufficient evidence that the population variance is larger in the old thermostat temperature readings. Fail to reject the null hypothesis, there is insufficient evidence that the population variance is larger in the old thermostat temperature readings.Reject the null hypothesis, there is insufficient evidence that the population variance is larger in the old thermostat temperature readings.
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