91影视

Given:

\({\rm{n = 83}}\)

\(\begin{array}{l}{\rm{33}}{\rm{.6,33}}{\rm{.8,33}}{\rm{.9,34}}{\rm{.1,34}}{\rm{.2,34}}{\rm{.6,34}}{\rm{.6,35,35}}{\rm{.2,35}}{\rm{.2,35}}{\rm{.4,35}}{\rm{.4,35}}{\rm{.4,}}\\{\rm{35}}{\rm{.5,35}}{\rm{.7,35}}{\rm{.8,36,30,36,36}}{\rm{.1,36}}{\rm{.1,30}}{\rm{.2,36}}{\rm{.4,30}}{\rm{.6,37,37}}{\rm{.4,37}}{\rm{.5,37}}{\rm{.5,38,}}\\{\rm{38}}{\rm{.7,38}}{\rm{.8,39}}{\rm{.8,41,42,42}}{\rm{.1,44}}{\rm{.6,48}}{\rm{.3,55}}\end{array}\)

The minimum is\(22.4\)

Since the number of data values is odd, the median is the middle value of the sorted data set:

\({\rm{M - }}{{\rm{Q}}_{\rm{2}}}{\rm{ - 33}}{\rm{.5}}\)

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

\({{\rm{Q}}_{\rm{1}}}{\rm{ - 30}}{\rm{.4}}\)

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

\({{\rm{Q}}_{\rm{3}}}{\rm{ - 36}}\)

The maximum is 55 .

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 \(75\% \)

The distribution is skewed to the right (or positively skewed), because the box in the boxplot is to the left between the whiskers.

Hence Skewed to the right (or positively skewed)

(b)

Given:

\({\rm{n = 83}}\)

\(\begin{array}{l}33.0,33.8,33.9,34.1,34.2,34.6,34.6,35,35.2,35.2,35.4,\\35.4,35.4,35.5,35.7,35.8,36,36,36,30.1,36.1,30.2,30.4,36.6,37,37.4,\\37.5,37.5,38,38.7,38.8,39.8,41,42,42.1,44.6,48.3,55\end{array}\)

NORMAL PROBABILITY PLOT

The data values are on the horizontal axis and the standardized normal scores are on the vertical axis.\({\rm{\{ 1,2,3, \ldots }}{\rm{.,n\} }}\).

The smallest standardized score corresponds with the smallest data value, the second smallest standardized score corresponds with the second smallest data value, and so on.

If the pattern in the normal probability plot is roughly linear and does not contain strong curvature, then the distribution is approximately normal. impression depth is normally distributed, because the rest of the plot is roughly linear.

Central limit theorem: If the sample size is large,then the sampling distribution of the sample mean\({\rm{\bar x}}\)is approximately normal. distribution assumption needed in order to calculate a confidence interval or bound for the true average depth\({\rm{\mu }}\)

Hence Plausible No normal assumption needed.

Given:

\(\begin{array}{l}{\rm{n = 83 }}\\{\rm{c = 99\% = 0}}{\rm{.99}}\end{array}\)

\(\begin{array}{l}33.6,33.8,33.9,34.1,34.2,34.6,34.6,35,35.2,35.2,35.4,\\35.4,35.4,35.5,35.7,35.8,30,30,30,30.1,30.1,30.2,30.4,30.0,37,37.4,\\37.5,37.5,38,38.7,38.8,39.8,41,42,42.1,44.6,48.3,55\end{array}\)

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

LARGE-SAMPLE CONFIDENCE INTERVAL

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

\({\rm{\bar x - }}\frac{{{\rm{22}}{\rm{.4 + 23}}{\rm{.6 + 24}}{\rm{.0 + \ldots + 44}}{\rm{.6 + 48}}{\rm{.3 + 55}}{\rm{.0}}}}{{{\rm{83}}}}{\rm{ - }}\frac{{{\rm{2769}}{\rm{.7}}}}{{{\rm{83}}}}{\rm{\gg 33}}{\rm{.3699}}\)

The variance is the sum of squared deviations from the mean divided by\({\rm{n - 1}}\). The standard deviation is the square root of the variance:

\(\begin{array}{l}{\rm{s - }}\sqrt {\frac{{{{{\rm{(22}}{\rm{.4 - 33}}{\rm{.3699)}}}^{\rm{2}}}{\rm{ + \ldots }}{\rm{. + (55}}{\rm{.0 - 33}}{\rm{.3699}}{{\rm{)}}^{\rm{2}}}}}{{{\rm{43 - 1}}}}} {\rm{\gg 5}}{\rm{.268 }}\\{{\rm{z}}_{{\rm{\alpha /2}}}}{\rm{ - 2}}{\rm{.575}}\\\end{array}\)

The margin of error is then:

\({\rm{E - }}{{\rm{z}}_{{\rm{\alpha /2}}}}{\rm{ \times }}\frac{{\rm{s}}}{{\sqrt {\rm{n}} }}{\rm{ - 2}}{\rm{.575 \times }}\frac{{{\rm{5}}{\rm{.2683}}}}{{\sqrt {{\rm{83}}} }}{\rm{\gg 1}}{\rm{.4891}}\)

The boundaries of the confidence interval then become:

\(\begin{array}{l}{\rm{\bar x - E - 33}}{\rm{.3699 - 1}}{\rm{.4891 - 31}}{\rm{.8808}}\\{\rm{\bar x + E - 33}}{\rm{.3699 + 1}}{\rm{.4891 - 34}}{\rm{.8590}}\end{array}\)

Hence the boundaries of the confidence interval then become \((31.8808,34.8590)\)