Question

I take a sample of 50 Americans and I am interested in how much time they spend on social networks every day and I know that the sample mean is 3.1. I know that the reports say that Americans spend on average 3 hours with a standard deviation of 0.29 hours. Since from my sample I obtained a higher value, I am interested to test the hypothesis that Americans spend at least 3 hours on social networks every day at a 5% significance level.

Answer 1

Solution:-

n = 50, x = 3.10, u = 3.0, S.D = 0.29

State the hypotheses. The first step is to state the null hypothesis and an alternative hypothesis.

Null hypothesis: u > 3
Alternative hypothesis: u < 3.0

Note that these hypotheses constitute a one-tailed test. The null hypothesis will be rejected if the sample mean is too small.

Formulate an analysis plan. For this analysis, the significance level is 0.05. The test method is a one-sample t-test.

Analyze sample data. Using sample data, we compute the standard error (SE), degrees of freedom (DF), and the t statistic test statistic (t).

SE = s / sqrt(n)

S.E = 0.041
DF = n - 1

D.F = 49
t = (x - u) / SE

t = 2.44

where s is the standard deviation of the sample, x is the sample mean, u is the hypothesized population mean, and n is the sample size.

The observed sample mean produced a t statistic test statistic of 2.44.

Thus the P-value in this analysis is 0.991

Interpret results. Since the P-value (0.991) is greater than the significance level (0.05), we cannot reject the null hypothesis.

From the above test we have sufficient evidence in the favor of the claim that Americans spend at least 3 hours on social networks every day at a 5% significance level.