2.Twelve individuals are randomized to two groups of 6 each. One group is given

Question

2.Twelve individuals are randomized to two groups of 6 each. One group

is given two cups of coffee and the other a glass of water. Ten minutes later

they are asked to memorize 10 words and after 5 minutes to recall as

many of these words as possible. If the alternative hypothesis is that those

given coffee will recall more words, what conclusion should we reach for

a=.1 with the following data:

1. Coffee

# words recalled

2. Water

# words recaled

5

2

3

5

7

9

9

7

8

6

8

5

What is the two-sided 95% CI?

1. Coffee

# words recalled

2. Water

# words recaled

5

2

3

5

7

9

9

7

8

6

8

5

Explanation / Answer

Solution:

Here, we have to use two sample t test for population means assuming pooled variance or equal variances. The null and alternative hypothesis for this test is given as below:

Null hypothesis: H0: There is no significant difference in the average number of words recalled for two groups.

Alternative hypothesis: Ha: The average number of words recalled is more for coffee group than that of water group.

H0: µ1 = µ2 versus Ha: µ1 > µ2

This is a one tailed test. (Upper tailed or right tailed test)

From given data, we have

Level of significance = = 0.1

n1 = 6

n2 = 6

X1bar = 6.666667

X2bar = 5.666667

S1 = 2.250926

S2 = 2.33809

df = n1 + n2 – 2 = 6 + 6 – 2 = 10

Upper critical value = 1.3722

(by using t-table or excel)

Test statistic formula for pooled variance t test is given as below:

t = (X1bar – X2bar) / sqrt[Sp2*((1/n1)+(1/n2))]

Where Sp2 is pooled variance

Sp2 = [(n1 – 1)*S1^2 + (n2 – 1)*S2^2]/(n1 + n2 – 2)

Sp2 = [(6 – 1)* 2.250926^2 + (6 – 1)* 2.33809^2]/(6 + 6 – 2)

Sp2 = 5.2667

t = (6.666667 – 5.666667) / sqrt[5.2667*((1/6)+(1/6))]

t = 0.7547

P-value = 0.2339

= 0.1

P-value > = 0.1

So, we do not reject the null hypothesis that there is no significant difference in the average number of words recalled for two groups.

There is insufficient evidence to conclude that the average number of words recalled is more for coffee group than that of water group.

Now, we have to find 95% confidence interval for difference between two population means.

Confidence interval = (X1bar – X2bar) -/+ t*sqrt[Sp2*((1/n1)+(1/n2))]

Critical value = t = 2.2281

Confidence interval = (6.666667 – 5.666667) -/+ 2.2281*sqrt[5.2667*((1/6)+(1/6))]

Confidence interval = 1.00 -/+ 2.2281*sqrt[5.2667*((1/6)+(1/6))]

Confidence interval = 1.00 -/+ 2.9522

Lower limit = 1.00 - 2.9522 = -1.9522

Upper limit = 1.00 + 2.9522 =3.9522

Confidence interval = (-1.9522, 3.9522)

The value of difference ‘0’ is lies within this interval, so we do not reject the null hypothesis in this case.

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