MAT 107 Project: Statistical Analysis of Book Page Data and Inferences

This assignment is a statistical project analyzing the number of pages in a sample of 30 statistics books. The solution includes the construction of a grouped frequency distribution and a histogram to visualize the data. The analysis determines the distribution's skewness and non-normality. The solution calculates z-scores for different page counts, identifies the absence of outliers, and determines a 95% confidence interval for the mean number of pages. Furthermore, the assignment includes a hypothesis test to determine if the mean number of pages is greater than a specified value, providing the test statistic, p-value, and a conclusion based on the significance level. The analysis covers various statistical concepts such as frequency distributions, histograms, z-scores, confidence intervals, and hypothesis testing to draw inferences from the given data set.

Question 1
Frequency distribution
Question 2
435 to 510 510 to 585 585 to 660 660 to 735 735 to 810 810 to 885
0
1
2
3
4
5
6
7
8
Histogram
Number of pages found in book
Frequency
Question 3
Based on the above histogram, it is apparent that the shape is asymmetric owing to present of
right skew. As a result, the distribution would be non-normal.
1

Question 4
Question 5
Question 6
2

Question 7
Question 8
Any data value less than ¿¿−2 σ__or greater than _¿+2 σ __is to be considered as outliers.
Outliers = None
Question 9
Mean 624.93
Standard deviation 116.28
z value for Books less than 750 pages ?
Sample size 30
Z score 5.89
Z score = (x- mean) /(standard deviation / sqrt(sample size)
Z score = (750-624.93)/(116.28/sqrt(30)) = 5.89
Question 10
Mean 624.93
Standard deviation 116.28
z value for Books has 350 pages ?
Sample size 30
Z score -12.95
3

Z score = (x- mean) /(standard deviation / sqrt(sample size)
Z score = (350-624.93)/(116.28/sqrt(30)) = 5.89
Question 11
The z score for 350 mean indicates that the standard deviation is -12.95 below the mean.
Question 12
95% confidence interval
Mean 624.93
Standard deviation 116.28
z value for 95% confidence interval 1.96
Sample size 30
Standard error = standard deviation / sqrt (sample size) 116.28/sqrt(30)) =21.230
Margin of error = Z value * standard error = 1.96*21.230 = 41.61
Lower value of 95% confidence interval = Mean – Margin of error = 624.93 – 41.61 = 583.32
Upper value of 95% confidence interval = Mean + Margin of error = 624.93 – 41.61 = 666.54
95% confidence interval = [583.32 666.54]
Question 13
It can be said with 95% confidence that the mean pages of books in statistical book would lie
between 583.32 and 666.54.
Question 14
Alternative hypothesis H1: μ>560 Mean number of pages is greater than 560.
4

Null hypothesis H0:μ=560 Mean number of pages is not greater than 560.
Test statistics (Z score) = (x- mean) /(standard deviation / sqrt(sample size)
Z score = (560-624.93)/(116.28/sqrt(30)) = -3.06
The p value = 0.001107
Given significance level = 0.05
Here p value less than significance level and hence, reject null hypothesis and accept
alternative hypothesis. Therefore, it can be concluded that the mean number of pages in
book is higher than 560.
5