Posted: October 17th, 2013

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Question and Answer

**Question Two**

** **Take into consideration to sets of data that give in mean difference (Md) of 4.5. From a researcher’s stand there would be need to establish whether this difference of 4.5 was due to treatment. If this study was to entail independent measures design, then there is a possibility that the treatment participants exhibited differing characteristics. For instance, a mean difference of 4.5 in examination results can imply that the participants in the first group were more inherently capable compared to participants in the second group. However, with repeated measure design, we cannot consider this problem because the same group of participants has been considered in the treatment. The other benefit associated with repeated measure design maintains that since the variance and SS are calculated when determining the score difference and individual differences from the eliminated samples. (For instance, a major difference between to values of data D and E is eliminated once the difference is taken into consideration). Since the individual differences have been eliminated Σ and the variance are reduced considerably. This in turn increases the probability of arriving to a significant result.

**Question Four**

**A)** If the researcher applied the use of different subjects (an independent measures design), then the result amounts to 20 subjects (10 x 2) for those that took part in the experiment.

**B)** If the researcher chose to use the same subjects for both groups (repeated measure design), then the number of subjects to participate in this study will be ten.

**C)** If the researcher decides to use a matched subject design, then the number of subjects to participate in the experiment will be (10 x 2) 20 (different subjects for the two groups)

**Question Six**

**A)** Ha: μd ≠ 0 vs Ho: μd = 0

Df = n – 1

n= 25

= 25 – 1 = 24

SE of the differences = s/√n

= 4/√25 = 4/5

t = (Md – μd)/SE = (4/5 – 0)/ 4/5 = 1

Critical t- score for α = 0.05 and Df = 24 is ± 2.3060

Since 1 < 2.3060, then we should not disregard Ho

In conclusion, that the result does not support the premise that this sample has not been acquired from a population that depicts μd equal to zero.

**B)** Ho: μd = 0 vs Ha: μd ≠ 0

Df = n – 1

n= 25

= 25 – 1 = 24

SE of the differences = s/√n = 12/√25 = 12/5

t = (Md – μd)/SE = (4 – 0)/1 = 4

Critical t- score for α = 0.05 and Df = 8 is ± 2.3060

Since 4 > 2.3060, we ca accept and reject Ha

The result suggest that the sample above has been derived from population where μd is equivalent to or greater than zero

**C)** Between the results (a) and (b) above, we are able to see that the standard has exhibited a decrease from 25 to 12. This implies less variability in the data hence more consistency in terms of treatment effect. This applies the same for the mean difference (Md of 4/5). In this case, the lesser the standard deviation becomes, the greater the probability of establishing the significance of the effect of treatment.

**Question 8**

**A)** Provided that n = 4, MD= 4, and s = 6, then freedom degrees for the repeated test measures can be obtained by the formula;

d.f.= n – 1

= 4-1 = 3. Consultation of the t distribution is conducted for a two tailed test where df = 3 and α=.05. The critical t values = + or – 3.182. Calculating the D scores variance will be done as follows. s2 = SS/n-1 = 64 (provided). Calculating the estimated standard for sample mean difference can be achieved using the following formula. Smd= √s2/n = √6/4 = 3/2. The repeated measures statistic t and can be calculated. The t statistic obtained is not from the critical region. Hence, the correct statistical decision involves failing to reject H0. This sample was acquired from a population with μd ≠ 0.

**B) **Provided that n =16, MD = 4 and s2 = 6, H0: μd = 0 (this implies that the sample was derived from a population where μd ≠ 0) H1: μd ≠ 0 (this implies that this sample was obtained from a population where μd ≠ 0). Therefore, freedom degrees for repeated measures can be obtained through the formula d.f. = n-1 = 16-1=15.

The t distribution can be consulted through a two tailed test where df = 15 and α = 0.05. The values for critical t are + or – 2.131. Estimated the standard error for the mean difference can be calculated through the following means: Smd = √s2/n = 6/6 = √1 =1. Hence, the repeated measures t statistic can be calculated:

t =Md – μd / Smd = 4-0 /2 = 2. The t statistic obtained comes from the critical region. Hence, it is credible to reject H0. This sample was achieved from μd ≠ 0 population.

C) The size of the sample poses an influence over the estimated standard error’s magnitude in the t statistic denominator. An increase in sample size translates to an increase in the value of t. hence, the probability of rejecting Ho increases as well.

**Question Ten**

**A)** (a) Ha: μd ≠ 0 vs Ho: μd = 0

Df = n – 1 = 25 – 1 = 24

SE of the differences = s/√n = 8/√25 = 1.6

t = (Md – μd)/SE = (4.7 – 0)/1.6 = 2.9375

Critical t- score for α = 0.05 and Df = 24 is ± 2.0639

Since 2.9375 > 2.0639, we reject Ho and accept Ha

**B) **Cohen’s d = Md/s = 4.7/8 = 0.5875

r^2 = t^2 / (t^2 + df) = 2.9375^2 / (2.9375^2 + 24) = 0.2645 (26.45%)

**Question 12**

**A)** I would feel depressed and surprised to doing worse than expected in an exam.

Ha: μd ≠ 0 vs Ho: μd = 0

Critical t- score for α = 0.05

t = (Md – μd)/SE = (1.28 – 0)/1.5 = 0.9375

This result is not sufficient to conclude that there is a significant difference in the ratings for self versus others.

**B)** Critical t- score for α = 0.9375, Critical t- score for α = 0.05,

t = (Md – μd)/SE = (1.28 – 0)/1.5 = 2.9375

hence we can calculate r2.

r2 0.9375 – 0.05 = 0.8875

**C) **The outcome of the hypothesis suggests most students achieve test results they expected.

**Question 14**

**A) **Provided that n = 6, MD= 7.4, and SS = 1215, then freedom degrees for the repeated test measures can be obtained by the formula;

d.f.= n – 1

= 6-1 = 5. Consultation of the t distribution is conducted for a two tailed test where df = 5 and α=.05. Effective cognitive performance variance = s2 = SS/n-1 = 64 (provided). Calculating the estimated standard for sample mean difference can be achieved using the following formula. Smd= √s2/n = √6/5 = 0.2784.Hence, this result supports the conclusion that the antioxidant supplement has a significant effect on cognitive performance.

**B)** 95 percent confidence interval = 6+/- 7.4 * SQRT [ 0.20 * (1 – 0.2784) / 1215] = [0.161 , 0.239] (16.1%, 23.9%)

**Question 16**

**A)** Provided that n = 9, MD= 16, and SS = 538, then freedom degrees for the repeated test measures can be obtained by the formula;

d.f.= n – 1

= 9-1 = 8. Consultation of the t distribution is conducted for a two tailed test where df = 30 and performance variance = s2 = SS/n-1 =538/ 8 = 67.25. Calculating the estimated standard for sample mean difference can be achieved using the following formula. Smd= √s2/n = √67.25/9 = 2.1374

**B) **df = Md/n = 16/8 = 2

r2 = t / df = 2.1374/2 = 1.067

**C) **The outcome of the hypothesis suggests that eating oatmeal regularly has health benefits

**Question 18**

**A)** Ho: μd = 0 vs Ha: μd ≠ 0

Df = n – 1 = 6 – 1 = 7

SE of the differences = s/√n = 12/√9 = 4

t = (Md – M)/Ss = (12 – 0)/4 = 3

Critical t- score for α = 0.05 and Df = 3 ± 2.7246

**B) **Smd = √s2/n = 6/6 = √1 =1. Hence, the variance for the sample of difference scoresand the estimated standard error for the mean difference can be calculated

t =12 – 0= 12

Smd = 4/0.05 /2 = 80

**Question 20**

**A)** H_{0}: (μ_{1} – μ_{2}) = 0

The alternative hypothesis says that there is a difference between the two population means.

H_{1}: (μ_{1} – μ_{2}) ≠ 0

Hence, α = .05 will be used

**B) **95 percent confidence interval = 29 +/- 20 * SQRT [ 0.20 * (13 – 1.7236) / 134] = [3.957 , 2.5074] (25.7%, 34.9%)

**Question 22**

The data provided was sufficient for the hypothesis: Rethinking or changing answers can significantly improve exam scores.

n = 22;

MD=2.5 points

s=3.1

α = .01

90 percent confidence interval = 22 +/- 1 * SQRT [ 0.25 * (12 – 2.5) / 22] = [1.947 , 2.8064] (28.9%, 32.7%)

**Question 24**

**A) **H_{0}: (μ_{1} – μ_{2}) = 0 vs H_{1}: (μ_{1} – μ_{2}) ≠ 0

Hence, α = .05 will be used since the alternative hypothesis says that there is a difference between the two conditions.

**B) **s2 = SS/n-1 = 117

Smd= √s2/n = √6/4 = 3/2

**C) **It appears that most people will feel relived by swearing in case of a straneous situation

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