Question

A researcher wants to examine the relationship between time spend on social media (variable X) and...

A researcher wants to examine the relationship between time spend on social media (variable X) and loneliness (variable Y) in young adults. A randomly sample of n = 72 young adults was asked how much time in average they spend on social media each day and how lonely day feel on a typical day.

The partial computations of collected data produced the following results:

                                            SP = 3.5     MX = 5   SSx = 16    MY = 20    SSY= 9

A. Based on these results, is there a significant correlation between time spend on social media and loneliness among young adults? Use a Pearson correlation test with p < .05, 2-tails to answer this research question.

Follow the steps of hypothesis testing and insert your answers below. In your calculations, round all numbers to two decimal places to avoid rounding errors.

ANSWER

  • H0:
  • H1:
  • Computed Pearson r =
  • df for decision about H0:
  • Critical r-value used for decision about H0:
  • Decision about H0 (i.e., reject or fail to reject):
  • Conclusion (i.e., is there a significant correlation or not? If the correlation is significant, is it positive or negative? Make sure to follow the APA reporting format in your conclusion to avoid loosing any points, see examples on the lecture slides.)

B. Find the regression equation for predicting a person's loneliness (i.e., variable Y) from the average time spend on social media each day (variable X). In your answer, show all computational steps.

Homework Answers

Answer #1

(a) The hypothesis being tested is:

H0: = 0

Ha: 0

r = 3.5/√16*9 = 0.292

df = 72 - 2 = 70

Critical r-value = 0.232

Since 0.292 > 0.232, we can reject the null hypothesis.

Therefore, we can conclude that there is a significant correlation between time spend on social media and loneliness among young adults.

(b) b1 = 3.5/16 = 0.21875

bo = 20 - 0.21875*5 = 18.90625

The regression line is:

y = 18.90625 + 0.21875*x



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