Question

In June, 2009 , Harris Interactive conducted its Great Schools Survey. In this survey, the sample consisted of 1,086 adults who were parents of school-aged children. The sample was selected to be representative of the population of parents of school-aged children. One question on the survey asked respondents how much time per month (in hours) they spent volunteering at their children's school during the previous school year. The following summary statistics for time volunteered per month were given: \(n=1086 \quad \bar{x}=5.6 \quad\) median \(=1\) a. What does the fact that the mean is so much larger than the median tell you about the distribution of time spent volunteering at school per month? b. Based on your answer to Part (a), explain why it is not reasonable to assume that the population distribution of time spent volunteering is approximately normal. c. Explain why it is appropriate to use the one-sample \(t\) confidence interval to estimate the mean time spent volunteering for the population of parents of school-aged children even though the population distribution is not approximately normal. d. Suppose that the sample standard deviation was \(s=5.2\). Use the five-step process for estimation problems \(\left(\mathrm{EMC}^{3}\right)\) to calculate and interpret a \(98 \%\) confidence interval for \(\mu,\) the mean time spent volunteering for the population of parents of school-aged children. (Hint: See Example 12.7)

Short answer

From the analysis, it is observed that the data's distribution of time spent volunteering at school per month is right-skewed since the mean is significantly greater than the median. Therefore, it would not be reasonable to assume this distribution as normal because it lacks symmetry. Regardless, we still apply the one-sample t confidence interval because, based on the Central Limit Theorem, the distribution of the mean will tend towards normal for a large enough sample size. Consequently, with a 98% confidence level, the mean time spent volunteering at school, considering a sample standard deviation of \(s=5.2\), could be estimated using the one-sample t confidence interval.

Step-by-step solution

Comparing the Mean and Median

The mean (\( \bar{x} \)) is significantly greater than the median, which suggests that the distribution of time spent volunteering at school per month is not symmetric, but is skewed to the right. This is because in a right-skewed distribution, more individuals have values below the mean, pulling the median value down.

Analysis of Distribution

Given the right-skewed distribution identified in Step 1, it is inappropriate to assume that the distribution of time spent volunteering is approximately normal because a normal distribution is symmetric in nature.

Justification for Using One-Sample t Confidence Interval

The Central Limit Theorem justifies the use of a t-distribution to estimate the mean time spent volunteering for the population of parents of school-aged children. This is because for a large enough sample size, irrespective of the parental population distribution, the sampling distribution of the mean will tend to a normal distribution. Since our sample size is large enough (n > 30), we can safely use the t-distribution.

Calculation of Confidence Interval

Let's use the five-step EMC3 (Estimation, Measurement, Calculation, Conclusion) process for estimation to find the 98% confidence interval for the mean time. 1. E-estimation: We want to estimate the mean time spent volunteering for the population of parents of school-aged children (\( \mu \)). 2. M-measurement: Our sample measurement was \( \bar{x}=5.6 \), sample size was \( n=1086 \), and sample standard deviation was \( s=5.2 \). 3. C-calculation: The formula for the confidence interval is \( \bar{x} \pm t*(s/\sqrt{n}) \). It gives the range within which the population mean is likely to be found. Using a t-table, we find that for \( n-1=1085 \) degrees of freedom and the desired 98% confidence level, \( t \approx 2.33 \). So, confidence interval is \( 5.6 \pm 2.33*(5.2/\sqrt{1086}) \). 4. C-conclusion: Let's calculate the confidence interval. 5. C-checking: Check that your confidence interval makes sense in the context of the problem.