Question

please answer quick and ill give good rating

2) An undergraduate psychology student has been told that the hours she spends readying for the verbal part of the GRE (a standardized test) will likely help predict how she will do on the exam. Her advisor tells her that the more hours she spends studying, the higher her grade will probably be. To test this, her advisor has comprised information from 10 psyc students in the past who have also taken the GRE and submitted the amount of study hours along with their score. Hopeful Grad Student Hours of Study GRE Score 10 280 300 287 270 270 2 3 4 5 6 7 8 9 10 45 33 15 25 20 95 65 20 30 260 337 300 270 329 a. What is the independent variable? b. What is the dependent variable? c. What test did you use to analyze the data? d. Is the test significant? How do you know? e. Write the statistical notation in APA style f. Report the findings in 1-3 sentences. Make sure to include the independent variable, the dependent variable, all relevant statistics and a summary.

Answer 1

Answer:-

Given that:-

(a)Independent variable$\rightarrow$

Hours of study

(b) Dependent Variable$\rightarrow$

GRE Score

(c) Correlation coefficient karl pearson's "correlation coefficient r" and t - test to analyze the data or Critical value of r test

(d)

x	y	$(x-\overline{x})^{2}$	$(y-\overline{y})^{2}$	$(x-\overline{x})(y-\overline{y})$
10	280	665.64	106.09	70617.7476
45	300	84.64	94.09	7963.7776
33	287	7.84	10.89	85.3776
15	270	432.64	412.09	178286.6176
25	270	116.64	412.09	48066.1776
20	260	249.64	918.09	229191.9876
95	337	3504.64	2180.89	7643234.3296
65	300	852.64	94.09	80224.8976
20	270	249.64	412.09	102874.1476
30	329	33.64	1497.69	50382.2916
358	2903	6197.60	6138.10	4628.60

$\overline{x}=\frac{358}{10}\overline{y}=\frac{2903}{10}$

$=35.8$$=290.3$ (using calculated)

correlation coefficient

$r=\frac{\sum (x-\overline{x})(y-\overline{y})}{\sqrt{\sum (x-\overline{x})^{2}}\sqrt{\sum (y-\overline{y})^{2}}}$

$r=\frac{4628.6}{\sqrt{6197.6}\sqrt{6138.1}}=\frac{4628.6}{6167.7783}$$=0.7505$

$t=\frac{r\sqrt{n-2}}{\sqrt{1-r^{2}}}=\frac{0.7505\sqrt{10-2}}{\sqrt{1-(0.7505)^{2}}}$

$=\frac{2.1227}{\sqrt{0.43675}}=\frac{2.1227}{0.661}=3.211$

P - value 0.012406 (d.of 2 two tailed test 0.05 significance

$H_{0}:P=0$

$H_{1}:P\neq 0$

p - value 0.012406<0.05

hence H_{0} Reject & H_{1} accepted.

There is sufficient evidence to conclude that there is a significant linear relationship between No.of hours of study and GRE score.

Note :- one another test is to check critical value of r from table critical value of r with degree of freedom (10-2) is 0.632

now calculated r(0.7504)>t_{table}

Hence result is siginificant.

sufficient evidence that there is a significant linear relation between hours of study and GRE score.

(e) Statistical notation in APA style

Mean (sample mdn

Mean square MS

mean square error MSE

\mu - sample mean, standard deviation

n- sample size population= \sigma

r- correlation Sample -s

standard error SEM(SE)

f) Report already given above

SEM is difference between predicted value and actual value and its mean is SEM.