Question

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1. Complete Part I of this assignment first, if you have not done so, because it provides guided practice on questions similar to the questions in Part II here.
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Q1) One-Sample t-Test (21 points total)

I am curious about how many hours per week my students spend on their online coursework. So, I surveyed 9 of my students on how many hours per week they spend on my online PSY 230 course. According to a survey of online college students (in all courses), students studied an average of 9 hours a week for a 3-credit course. I assume that the national study time scores are normally distributed, and I set the significance level at α = .05.

Student	Study hours
1	12
2	19
3	21
4	14
5	11
6	9
7	12
8	10
9	7

1. What is the dependent variable in this hypothesis test? What is the “sample” for this one-sample t test? (2 points total: 1 for each question)

2. I am interested in testing whether there are any differences between my students and the national average, so my hypothesis is non-directional. What would be the null and alternative hypotheses in both words and symbol notations for my analysis? (4 points total: 1 for each hypothesis in words, 1 for each hypothesis in notation)

3. Calculate the sample mean. (2 points total: 1 for work, 1 for answer)

4. Estimate the standard deviation of the population (2 points total: 1 for work, 1 for answer)

5. Calculate the standard error (standard deviation of the sampling distribution) (2 points total: 1 for work, 1 for answer)

6. Calculate the t statistic for the sample (2 points total: 1 for work, 1 for answer)

7. Specify whether the test is a one-tailed or two-tailed test based on the hypotheses form (b), and then determine the critical t value(s) based on the type of test and the preset alpha level (2 points total: 1 for each answer)

8. Compare the t statistic with the critical t value. Is the calculated t statistic more extreme or less extreme than the critical t value? Then make a decision by stating whether we “reject” or “fail to reject” the null hypothesis (2 points total: 1 for each answer)

1. Interpret the result in 1-2 sentences (you may restate the hypothesis accepted or explain it in your own words) (1 point)

10. Calculate the raw effect size and the standardized effect size of this hypothesis test (4 points total: 2 per effect size: 1 for work, 1 for answer)

Answer 1

One sample t test is conducted when you have a known values for your sample data.In this you might want to compare the population mean with the sample mean you have, and you dont know the population variance .or in the case you have a small sample

a) In a testing of hypothesis the dependent variable is the one which is being tested or neasured , here we are studying about the study time so the dependent variable is the study time .

Here the sample is the 9 study time score data of 9 students collected .

b) The Null hypothesis H0: There is no significant difference between the population mean and the sample mean

H0:

and the alternative hypothesis will be H1: There is difference between the population mean and the sample mean

H1:

c) To calculate sample mean we ahve the formula

Here and n= 9 so = sample mean = 12.77

d) We have the formula for estimating the population SD as

Where is the population standard devation and = 9 hours (given)

So we have the Population SD=

X	X-
12	3	9
19	10	100
21	12	144
14	5	25
11	2	4
9	0	0
12	3	9
10	1	1
7	-2	4

Sum= 296/9=32.88

Square root of 32.88 = 5.73 which is the estimated value of population SD

e) Standard error is the standard devation of the sample

in the previous formula we need to cange the population mean as sample mean and the equation will become

The Standard error

x	x-xbar
12	-.7	.49
19	6.3	39.69
21	8.3	68.89
14	1.3	1.69
11	-1.7	2.89
9	-3.7	13.69
12	-.7	.49
10	-2.7	7.29
7	-5.7	32.49

Sum =167.61/9=18.62 Square root = S=4.31

f) from b we have taken the null hypothesis as Population mean  equal smaple mean and the alternate hypothsis are population mean not equal to the sample mean to it is a two tailed test , if we would have taken greater than or less than we would have got a one tailed test

g) The test statistic will be = 12.77-9/4.31/3=3X3/4.31=9/4.31=2.088

h) since we have the 2 tailed test we have p value is the probability that the t statistic  with degrees of freedom 8 (n-1) is less than -2.088 or greater than +2.088

From the t distribution calculator applying the values of T score that is 2.088, Degree of freedom =8 two tailed and also significance level =.05

P value = .0702> .05

If we have the pa value greater than the significance level we need to reject the null hypothesis

So we have to conclude that there is significance difference between the mena value of study time of the sample and national average

Effect size