91影视

(a) Given:

\(\begin{array}{l}{\rm{n = 48 }}\\{\rm{c - 95\% = 0}}{\rm{.95}}\end{array}\)

\(\begin{array}{l}62,50,53,57,41,53,55,61,59,64,50,53,64,62,50\\,68,54,55,57,50,55,50,50,55,40,55,53,54,52,47,\\47,55,57,48,03,57,57,55,53,59,53,52,50,55,60,50,50,58\end{array}\)

Sort the data values from smallest to largest:

\(\begin{array}{l}41,40,47,47,48,50,50,50,50,50,50,50,52,52,53,53,53,53,53,53,54\\,54,55,55,55,55,55,55,55,55,50,50,57,\\57,57,57,57,58,59,59,60,61,02,02,63,04,64,68\end{array}\)

The minimum is \(41\) .

Since the number of data values is even, the median is the average of the two middle values of the sorted data set:

\({\rm{M - }}{{\rm{Q}}_{\rm{2}}}{\rm{ - }}\frac{{{\rm{55 + 55}}}}{{\rm{2}}}{\rm{ - 55}}\)

The first quartile is the median of the data values below the median (or at 25 \% of the data):

\({{\rm{Q}}_{\rm{1}}}{\rm{ - }}\frac{{{\rm{50 + 52}}}}{{\rm{2}}}{\rm{ - 51}}\)

The third quartile is the median of the data values above the median (or at 75 \% of the data):

\({{\rm{Q}}_{\rm{3}}}{\rm{ - }}\frac{{{\rm{57 + 57}}}}{{\rm{2}}}{\rm{ - 57}}\)

The maximum is \(08\) .

BOXPLOT

The whiskers of the boxplot are at the minimum and maximum value. The box starts at the first quartile, ends at the third quartile and has a vertical line at the median.

The first quartile is at \(25\% \)of the sorted data list, the median at \(50\% \)and the third quartile at \(75\% \).

Hence The distribution appear to be roughly symmetric, because the box of the boxplot lies roughly in the middle between the whiskers.

(b) Central limit theorem: If the sample size is large (more than 30), then the sampling distribution of the sample mean \bar{x} is approximately normal.

LARGE-SAMPLE CONFIDENCE INTERVAL

The sample mean is the sum of all data values divided by the sample size:

\({\rm{\bar x - }}\frac{{{\rm{62 + 50 + 53 + \ldots + 50 + 56 + 58}}}}{{{\rm{48}}}}{\rm{ - }}\frac{{{\rm{2626}}}}{{{\rm{48}}}}{\rm{\gg 54}}{\rm{.7083}}\)

The variance is the sum of squared deviations from the mean divided by n-1. The standard deviation is the square root of the variance:

\({\rm{s - }}\sqrt {\frac{{{{{\rm{(62 - 54}}{\rm{.7083)}}}^{\rm{2}}}{\rm{ + \ldots }}{\rm{. + (58 - 54}}{\rm{.7083}}{{\rm{)}}^{\rm{2}}}}}{{{\rm{48 - 1}}}}} {\rm{\gg 5}}{\rm{.2307}}\)

\({{\rm{z}}_{{\rm{\alpha /2}}}}{\rm{ - 1}}{\rm{.96}}\)

The margin of error is then:

\({\rm{E - }}{{\rm{z}}_{{\rm{\alpha /2}}}}{\rm{ \times }}\frac{{\rm{s}}}{{\sqrt {\rm{n}} }}{\rm{ - 1}}{\rm{.96 \times }}\frac{{{\rm{5}}{\rm{.2307}}}}{{\sqrt {{\rm{48}}} }}{\rm{\gg 1}}{\rm{.4798}}\)

The boundaries of the confidence interval then become:

\(\begin{array}{l}{\rm{\bar x - E - 54}}{\rm{.7083 - 1}}{\rm{.4798 - 53}}{\rm{.2285}}\\{\rm{\bar x + E - 54}}{\rm{.7083 + 1}}{\rm{.4798 - 56}}{\rm{.1881}}\end{array}\)

Hence The boundaries of the confidence interval then become is \((53.2285,56.1881)\)

(c) Given:

\(\begin{array}{l}{\rm{E = 1}}\\{\rm{c = 95\% = 0}}{\rm{.9540}}\\{\rm{ to 70}}\end{array}\)

The population standard deviation can be estimated by one forth of the range of values:

\({\rm{\sigma \gg }}\frac{{{\rm{70 - 40}}}}{{\rm{4}}}{\rm{ - }}\frac{{{\rm{30}}}}{{\rm{4}}}{\rm{ - 7}}{\rm{.5}}\)

Formula sample size:

\(\begin{array}{l}{\rm{n - }}{\left( {\frac{{{{\rm{z}}_{{\rm{\alpha /2}}}}{\rm{\sigma }}}}{{\rm{E}}}} \right)^{\rm{2}}}\\{{\rm{z}}_{{\rm{\alpha /2}}}}{\rm{ - 1}}{\rm{.96}}\end{array}\)

The sample size is then (round up to the nearest integer!):

\({\rm{n - }}{\left( {\frac{{{{\rm{z}}_{{\rm{\alpha /2}}}}{\rm{\sigma }}}}{{\rm{E}}}} \right)^{\rm{2}}}{\rm{ - }}{\left( {\frac{{{\rm{1}}{\rm{.96 \times 7}}{\rm{.5}}}}{{\rm{1}}}} \right)^{\rm{2}}}{\rm{\gg 217}}\)

Hence the sample size is \({\rm{n = 217}}\)