Question

Increased social media usage is being linked to fewer interpersonal relations and increased loneliness across today’s younger generation. On average, millennials spend 6 hours per week on social media. Connect-U is a new business which attempts to connect people via social media while also encouraging in-person interactions. Connect-U is hoping to decrease the effects of social media usage on loneliness by decreasing total time spent using social media per week. After 6 months of operations, the average time a sample of 100 customers spent using social media per week was 5.75 hours. Historical records indicate that the standard deviation for weekly social media usage is 1 hour. You would like to perform a test to determine whether Connect-U is successful at its mission.

Based on this test, the correct decision at a 95% confidence level would be:

a. Fail to reject H₀, conclude that the new business does reduce social media usage

b. Reject H₀, conclude the new business does not reduce social media usage

c. Fail to reject H₀, conclude that the new business does not reduce social media usage

d. Reject H₀, conclude the new business does reduce social media usage

Answer 1

sample mean, xbar = 5.75
sample standard deviation, s = 1
sample size, n = 100
degrees of freedom, df = n - 1 = 99

Given CI level is 95%, hence α = 1 - 0.95 = 0.05
α/2 = 0.05/2 = 0.025, tc = t(α/2, df) = 1.984

ME = tc * s/sqrt(n)
ME = 1.984 * 1/sqrt(100)
ME = 0.198

CI = (xbar - tc * s/sqrt(n) , xbar + tc * s/sqrt(n))
CI = (5.75 - 1.984 * 1/sqrt(100) , 5.75 + 1.984 * 1/sqrt(100))
CI = (5.55 , 5.95)

d. Reject H0, conclude the new business does reduce social media usage

because confidence interval does not contain 6

Below are the null and alternative Hypothesis,
Null Hypothesis, H0: μ = 6
Alternative Hypothesis, Ha: μ < 6

Test statistic,
t = (xbar - mu)/(s/sqrt(n))
t = (5.75 - 6)/(1/sqrt(100))
t = -2.5

P-value Approach
P-value = 0.007
As P-value < 0.05, reject the null hypothesis.