Q2-20BSC
Question
Years in college Listed below are the numbers of years it took for a random sample of college students to earn bachelor’s degrees (based on the data from the National Center for Education Statistics). Construct a 95% confidence interval estimate of the mean time for all college students to earn bachelor’s degrees. Does it appear that college students typically earn bachelor’s degrees in four years? Is there anything about the data that would suggest that the confidence interval might not be good result?
4 4 4 4 4 4 4.5 4.5 4.5 4.5 4.5 4.5
6 6 8 9 9 13 13 15
Step-by-Step Solution
VerifiedThe \({\rm{95\% }}\) confidence interval is from 4.86 years to 8.14 years.
No, it does not appear that college students typically earn bachelor’s degrees in four years.
The confidence interval method is a good result.
The number of years it took for 20 college students to earn a bachelor’s degree is recorded. The confidence interval is \(95\% \) .
Let \(n = 20\) be the number of selected students.
The mean value is given below:
\(\begin{array}{c}\bar x = \frac{{\sum\limits_{i = 1}^n {{x_i}} }}{n}\\ = \frac{{4 + 4 + .... + 13 + 15}}{{20}}\\ = \frac{{134}}{{20}}\\ = 6.5\end{array}\)
The mean value is 6.5.
The standard deviation of the given students is:
\(\begin{array}{c}{s_{}} = \sqrt {\frac{{\sum\limits_{i = 1}^n {{{({x_i} - \bar x)}^2}} }}{{n - 1}}} \\ = \sqrt {\frac{{{{\left( {4 - 6.5} \right)}^2} + {{\left( {4 - 6.5} \right)}^2} + ... + {{\left( {13 - 6.5} \right)}^2} + {{\left( {15 - 6.5} \right)}^2}}}{{20 - 1}}} \\ = 3.506\end{array}\)
Therefore, the standard deviation is 3.506.
Since the value of the sample size is lesser than 30 and the population standard deviation is unknown, use t-distribution.
The degrees of freedom are as follows:
\(\begin{array}{c}{\rm{df}} = n - 1\\ = 20 - 1\\ = 19\end{array}\)
At \(95\% \) confidence interval and \(\alpha = 0.05\) with 20 degrees of freedom, the critical value can be obtained using the t-table.
\(\begin{array}{c}{t_{crit}} = {t_{\frac{\alpha }{2},df}}\\ = {t_{\frac{{0.05}}{2},19}}\\ = 2.09\end{array}\)
The margin of error is computed as follows:
\(\begin{array}{c}E = {t_{crit}} \times \frac{s}{{\sqrt n }}\\ = 2.09 \times \frac{{3.506}}{{\sqrt {20} }}\\ = 1.6407\end{array}\)
The formula for the confidence interval is given as follows:
\(\begin{array}{c}CI = \bar x - E < \mu < \bar x + E\\ = \left( {6.5 - 1.6407 < \mu < 6.5 + 1.6407} \right)\\ = (4.86 < \mu < 8.14)\end{array}\)
This means there is a 95% confidence that the mean time required for all college students to earn a bachelor's degree lies between (4.86 years and 8.14 years).
Since the value of four years does not fall between the 95% range estimate of 4.86 and 8.14, it cannot be inferred that college students typically earn a bachelor’s degree in four years.
The data is independent and randomly chosen, but most sample points do not lie in the range estimate. Hence, the confidence interval is not a good result.