Question

It may be that sunshine has a unique effect on learning statistics. Previous research has been in disagreement, with some studies showing that sunshine increases amount learned whereas others show sunshine has detrimental effects on learning. You would like to determine for yourself whether or not sunshine makes a DIFFERENCE on statistics learning. Let's assume you take five statistics students and give them a lesson on a sunny day, and then take the same students and give them the related lesson on a rainy day. These are their results for a quiz on their lesson:

Sunny Day Results:

• Student 1: 10.8
• Student 2: 7.9
• Student 3: 8.3
• Student 4: 9.2
• Student 5: 6.1

Rainy Day Results:

• Student 1: 9.2
• Student 2: 7.7
• Student 3: 7.8
• Student 4: 10.5
• Student 5: 8.0

a. State the null and alternative hypotheses

b. What are the degrees of freedom for your t test? Find the corresponding critical t-value for Type I error rate (alpha) of α = 0.05.

c. Calculate your observed t-statistic (hint: you will need to calculate the average difference scores and the SD of the average difference score first).

d. Compare your observed t-statistic to the critical t-value(s). What do you conclude regarding the null hypothesis?

e. Calculate and interpret the 95% Confidence interval.

f. Calculate and interpret the standardized effect size (Cohen's d).

g. What do you conclude about your research question (use your own words, in everyday language).

Answer 1

a)

Let us denote population mean score for sunny day and rainy day as $\mu _{1}$ and $\mu _{2}$ respectively.

We are testing a claim whether these two means are equal or not.

Therefore,

Null hypothesis:

$H_{0}:\mu _{1}=\mu _{2}$

Alternative hypothesis:

$H_{1}:\mu _{1}\neq \mu _{2}$

b) Degrees of freedom = n - 1 = 5 - 1 = 4

Level of significance:

= 0,05

Note that this is a two tailed test.

Therefore in order to find t critical we have to consider probbaility = $\alpha/2$

$\therefore \alpha/2=\frac{0.05}{2}=0.025$

t-critical:

Similarly

Region of acceptance of null hypothesis is between t = -2.776 to t = 2.776

c) t- statistic:

	Sunny day	Rainy day	Difference
	10.8	9.2	1.6
	7.9	7.7	0.2
	8.3	7.8	0.5
	9.2	10.5	-1.3
	6.1	8	-1.9
mean	8.46	8.64	-0.18
SD	1.727136358	1.20124935	1.413152504
VAR	2.983	1.443	1.997

$t=\frac{\bar{d}}{\frac{s_{d}}{\sqrt{n}}}=\frac{-0.18}{\frac{1.413152504}{\sqrt{5}}}=-0.1138$

d)

We see that

Test statistic value falls in the region of accceptance.

Hence we fail to reject null hypothesis.

e) 95% confidence interval:

Confidence interval is given by the formula:

$\bar{d}\pm t*\frac{s_{d}}{\sqrt{n}}$

Plugging in values we get,

$-0.18\pm 2.227*0.6319810124=-0.18\pm1.75437929$

$\Rightarrow CI=(-1.9344,1.5744)$

That means we are 95% sure that population mean difference score falls within -1.9344 and 1.5744

f) Cohen's d formula:

$Cohen'sd=\frac{m_{1}-m_{2}}{\sqrt{\frac{var_{1}+var_{2}}{2}}}=\frac{8.46-8.64}{\sqrt{\frac{2.983+1.443}{2}}}$

$\Rightarrow Cohen'sd=\frac{-0.18}{1.487615542}=-0.121$

Here value of Cohen's d represent small decreacring effect.

g) We conclude that sunshine has no effect on learning thus we may assume that learning capacity in both sunny and rainy days is same.