Question

Q2) Paired-Samples t Test (14 points) A research project has been tracking the health and cognitive functions of the elderly population in Arizona. The table below shows the memory test scores from 10 elderly residents, tested first when they were 65 years old and again when they were 75 years old. The researcher wants to know if there is a significant decline in memory functions from age 65 to age 75 based on this sample. In other words, it is hypothesized that the memory score at age 75 is significantly lower than the memory score at age 65. So the null and alternative hypotheses should be directional. The alpha level was set at a = .05 for a one- tailed hypothesis test Memory score Subject Age 65 Age 75 1 62 60 95 88 2 55 56 90 89 98 90 73 75 73 70 71 75 82 80 10 66 62 c. What would be the null and alternative hypotheses in both words and symbol notations? (2 points total: 1 for each hypothesis, .5 for written and .5 for notation) d. Calculate the difference score by subtracting each "Age 65" score from the associated "Age 75" score for each subject. Fill in the column in the table below for "difference score." (1point total: deduct .5 for each error up to 1 point.) g. Calculate the standard error (standard deviation of the sampling distribution) (1 point total: .5 if process is correct but answer is wrong) h. Calculate the t statistic for the sample of difference scores (1 point total: .5 if process is correct but answer is wrong) į. Figure out the degree of freedom, and then determine the critical t value(s) based on the type of test and the preset alpha level. (1point total: .5 for df, .5 for critical t value) j. Compare the t statistic with the critical t value. Is the calculated t statistic more extreme or less extreme than the criticalt value? Then make a decision about the hypothesis test, stating explicitly "reject" or "fail to reject" accordingly. (2 points total: 1 for each answer) k. Interpret the result in 1-2 sentences to answer the research question (you may use the wording from the hypothesis or explain it in your own words) (1 point) I.Calculate the standardized effect size of this hypothesis test (1 point: .5 if process is correct but answer is wrong)

Answer 1

SOLUTION:

PART c)

Null hypotheses and Alternative hypotheses in words:

It is assumed that there is no decline in the memory, hence,

Null hypotheses: The memory score at age 65 is not greater than the memory score at age 75.

We have to find out if the memory score at age 75 is lower than the memory score at the age 65. Hence,

Alternative hypotheses: The memory score at age 75 is significantly lower than the memory score at age 65 or the memory score at the age 65 is significantly greater than at the age 75.

Null hypotheses and Alternative hypotheses in symbols:

If $\mu$_d is the mean difference between the memory scores at the age 65 and age 75 then;

Null hypotheses: H₀ :   $\mu$_d =0

Alternative hypotheses: H_a : $\mu$_d >0

PART d):

	memory score	memory score
Subject	Age 65	Age 75	difference score(age 65 - age 75) (d)
1	62	60	2
2	95	88	7
3	55	56	-1
4	90	89	1
5	98	90	8
6	73	75	-2
7	73	70	3
8	71	75	-4
9	82	80	2
10	66	62	4

PART g): Standard deviation of the sampling data i.e., difference score:

the formula for calculation of standard deviation for a sample of difference scores is:

s_d = $\sqrt{\frac{\Sigma d^{2}-\frac{1}{n}({\Sigma d})^{2}}{n-1}}$

where n is the no. of observations, i.e., n =10

calculating all the required values:

	memory score	memory score
Subject	Age 65	Age 75	difference score(age 65 - age 75) (d)	d2
1	62	60	2	4
2	95	88	7	49
3	55	56	-1	1
4	90	89	1	1
5	98	90	8	64
6	73	75	-2	4
7	73	70	3	9
8	71	75	-4	16
9	82	80	2	4
10	66	62	4	16
	M1= 76.5	M2 = 74.5	$\Sigma$d = 20	$\Sigma$d2 = 168

substituting all the values in the formula:

s_d = $\sqrt{\frac{168-\frac{1}{10}20^{2}}{10-1}}$ = $\sqrt{\frac{168-40}{9}}$ = 3.77

PART h): t statistic for the sample of difference score

the test statistic for a sample of pairs is given by:

T = $\frac{\bar{d}-D_{0}}{\frac{s_{d}}{\sqrt{n}}}$

where, $\bar{d}$ is the point estimate for the sample of difference scores calculated by the formula same as mean

$\bar{d}$=$\Sigma$d/n = 20/10 = 2

and D₀ can be taken from null hypotheses as Null hypotheses: H_{0 :}$\mu _{0}$ = D₀

Hence, in our question D₀ = 0 as null hypotheses is H_{0 :}$\mu _{0}$ = 0

Now, substituting all the values in the formula:

T = (20-0)/(3.77/$\sqrt{}$10)

T = 20/(3.77/3.16)

T = 20/1.19 = 16.81

PART i): degree of freedom and critical value(s):

The test statistic has Student’s t-distribution and degree of freedom is calculated as:

dof = n-1

where n is the size of the sample which is n=10

therefore, dof = 10-1 = 9

Since the symbol in H_a is “>” this is a right-tailed test, so there is a single critical value, _{$t_{\alpha }$}= t_0.05 with 9 degrees of freedom, which from the Critical Value table for t values we read off as 1.833.

PART j): compare the t statistic with critical t value:

since, this is a right tailed test and the critical value of t is 1.833, the rejection region will be [1.833,$\infty$).

here we see that, 1.833 < 16.81 < $\infty$

the test statistic is more extreme to the critical value.

i..e, the test statistic falls in the rejection region. Therefore, the decision is to reject H₀.

PART k):  In the context of the problem our conclusion is:

The data provide sufficient evidence, at the 5% level of significance, to conclude that the memory score at age 75 is significantly lower than the memory score at age 65 or the memory score at the age 65 is significantly greater than at the age 75.

PART l): Standardized effect size

the standardized effect size is calculated as:

ses = (M1 - M2)/ s_d

where M1 and M2 are means of memory scores at the age 65 and 75 respectively and s_d is the standard deviation of differences

therefore,

ses = (76.5-74.5)/3.77 = 0.53