## 1.2 Carrying out the test: One-sample test of proportions

We are now ready to carry out the one-sample test of proportions to test whether the proportion of social media users who use Facebook more than once per day is different from 0.73. The results of the test are as follows:


1-sample proportions test with continuity correction

data:  368 out of 484, null probability 0.73
X-squared = 2.1078, df = 1, p-value = 0.1466
alternative hypothesis: true p is not equal to 0.73
95 percent confidence interval:
0.7192645 0.7971860
sample estimates:
p
0.7603306 

We note the following:

• The test statistic is equal to 2.1078
• $$p$$-value is equal to 0.1466 Since this is larger than $$\alpha = 0.05$$, we cannot reject $$H_0$$.
• The 95% confidence interval for $$p$$ is (0.7193, 0.7972), meaning that we are 95% confident that the true value of $$p$$ lies within the interval (0.7193, 0.7972). Since $$p_0 = 0.73$$ is included in this interval, we cannot reject $$H_0$$ at the $$\alpha = 0.05$$ level of significance.
• The sample proportion is $$\hat{p} = 0.7603$$.