A researcher records the amount of time (in minutes) that parent child pairs spend on social networking sites to test whether they show any generational differences. From the following findings in APA format, interpret these results by stating the research design used (repeated measures or matched pairs), the sample size, the decision, and the effect size.

Parents spend significantly less time on social networking sites compared their children (MD = -42 minutes), t(29) = 4.021, p < .05, d = 0.49.

(a) What research design was used? (Repeated measures or matched paires?)

(b) What is the sample size? (n = ?)

(c) What is the decision? (Retain or reject the null?)

(d) Based on the estimated Cohen's d, d = 0.49, what is the effect size? (Medium, Large, Small, or Trivial?)

(a) A matched pair design is used to perform the test.

(b) Each sample is of size 30.

(c) The null hypothesis is rejected.

(d) The effect size of the test is medium.

Step-by-step explanation:

A statistical experiment is performed to determine whether the amount of time (in minutes) that parent-child pairs spend on social networking sites to test show any generational differences.

Given:

MD = -42 minutes

t₍₂₉₎ = 4.021

p < 0.05

d = 0.49

The hypothesis is:

H₀: There is no difference between the means, i.e. μ₁ = μ₂.

Hₐ: There is a significant difference between the means, i.e. μ₁ ≠ μ₂.

(a)

A matched pair design is used to perform the test.

In a matched pair design the participant of the two groups are related in some way. For example, the same sample is tested before and after applying the treatment.

In this case the matched are pairs are of children and their parents. And the treatment is the amount of time spent on social networking sites.

(b)

The degrees of freedom of a matched pair design is:

df = n - 1

29 = n - 1

n = 29 + 1

n = 30

Thus, each sample is of size 30.

(c)

The p-value of the test statistic is < 0.05.

If the p-value is less than the significance level α (say, 0.05) then the null hypothesis is rejected and vise-versa.

Since p < 0.05 the null hypothesis is rejected concluding that there is a significant difference between the amount of time spent on social networking sites by parents and children.

(d)

The estimated value of Cohen's d is, d = 0.49.

A Cohen's d value between the range 0.50 to 0.80 is considered as medium.

Thus, the effect size of the test is medium.

(sin 42)/31. = sinx)/37

sinx = 37 (sin42)/31

= 37× 0.66/31 = 0.787

x= 51.9

other angle opposite to y = 180-51.9-42

= 180- 93.9 =86.1

so

sin42)/31 = sin 86.1)/y

0.021 = 0.997 /y

y= 0.997/0.021 =47.47

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step-by-step explanation: