Question

Seventy-seven students at the University of Virginia were asked to keep a diary of a conversation with their mothers, recording any lies they told during this conversation (San Luis Obispo Telegram-Tribune, August 16, 1995). The mean number of lies per conversation was 0.5 . Suppose that the standard deviation (which was not reported) was 0.4 a. Suppose that this group of 77 is representative of the population of students at this university. Construct a \(95 \%\) confidence interval for the mean number of lies per conversation for this population. b. The interval in Part (a) does not include \(0 .\) Does this imply that all students lie to their mothers? Explain.

Short answer

The interval does not include 0, but this does not imply that all students lie to their mothers. The confidence interval instead provides a range of plausible values for the mean number of lies per conversation. It stating that with 95% confidence, the interval contains the true population mean number of lies which students tell their mothers during a conversation.

Step-by-step solution

Constructing the Confidence Interval

The formula for constructing a confidence interval is given by\[\bar{x} \pm z \left(\frac{s}{\sqrt{n}}\right)\]where \(\bar{x}\) is the sample mean, \(z\) is the z-score (which for a 95% confidence interval is approximately 1.96), \(s\) is the sample standard deviation, and \(n\) is the sample size. Here, \(\bar{x} = 0.5\), \(s = 0.4\), and \(n = 77\). Hence, the confidence interval will be:\[0.5 \pm 1.96(\frac{0.4}{\sqrt{77}})\]

Calculating the Confidence Interval

Now, substitute the given values into the confidence interval formula:\[0.5 \pm 1.96(\frac{0.4}{\sqrt{77}})\]and calculate the result. This will give the 95% confidence interval.

Interpreting the Confidence Interval

Based on the calculated confidence interval, interpret whether the interval includes 0 or not. If it does not, provide an explanation. Also provide an analysis of what it would mean if the interval does not contain 0.