Mean and Variance | Introduction to Statistics and Data Analysis | Probability & Statistics for Engineers & Scientists | Walpole | Problem 1.10

Solution:

For the company A:

We know, based on our answer in Exercise 1.4, that the sample mean for samples in Company A is $\displaystyle 7.950$.

To compute for the sample variance, we shall use the formula

$\displaystyle s^2=\sum _{i=1}^n\frac{\left(x_i-\overline{x}\right)^2}{n-1}$

The formula states that we need to get the sum of $\displaystyle \left(x_i-\overline{x}\right)^2$, so we can use a table to solve $\displaystyle \left(x_i-\overline{x}\right)^2$ for every sample.

Note: You can refer to the solution of Exercise 1.7 on how to use a table to solve for the variance.

The variance is

$\displaystyle \left(s^2\right)_A=\sum _{i=1}^{10}\:\frac{\left(x_i-7.950\right)^2}{10-1}$

$\displaystyle =1.2078$

The standard deviation is just the square root of the variance. That is

$\displaystyle s_A=\sqrt{1.2078}=1.099$

For Company B:

Using the same method employed for Company A, we can show that the variance and standard deviation for the samples in Company B are

$\displaystyle \left(s^2\right)_B=0.3249$ and $\displaystyle s_B=0.570$.

Variance & Standard Deviation | Introduction to Statistics and Data Analysis | Probability & Statistics for Engineers & Scientists | Walpole | Problem 1.9

Solution:

For the samples with NO AGING:

We know, based on our answer in Exercise 1.3, that the sample mean for samples with no aging is $\displaystyle \overline{x}_{no\:aging}=222.10$.

To compute for the sample variance, we shall use the formula

$\displaystyle s^2=\sum _{i=1}^n\frac{\left(x_i-\overline{x}\right)^2}{n-1}$

The formula states that we need to get the sum of $\displaystyle \left(x_i-\overline{x}\right)^2$, so we can use a table to solve $\displaystyle \left(x_i-\overline{x}\right)^2$ for every sample.

Note: You can refer to the solution of Exercise 1.7 on how to use a table to solve for the variance.

The variance is

$\displaystyle \left(s^2\right)_{no\:aging}=\sum _{i=1}^{10}\:\frac{\left(x_i-\overline{x}\right)^2}{n-1}$

$\displaystyle =\frac{1}{10-1}\left[\left(227-222.10\right)^2+\left(222-222.10\right)^2+...+\left(221-222.10\right)^2\right]$

$\displaystyle =42.12$

The standard deviation is just the square root of the variance. That is

$\displaystyle s_{no\:aging}=\sqrt{s^2}=\sqrt{42.12}=6.49$

For the samples with AGING:

We know, based on our answer in Exercise 1.3, that the sample mean for samples with aging is $\displaystyle \overline{x}_{no\:aging}=209.90$.

To compute for the sample variance, we shall use the formula

$\displaystyle s^2=\sum _{i=1}^n\frac{\left(x_i-\overline{x}\right)^2}{n-1}$

The formula states that we need to get the sum of $\displaystyle \left(x_i-\overline{x}\right)^2$, so we can use a table to solve $\displaystyle \left(x_i-\overline{x}\right)^2$ for every sample.

Note: You can refer to the solution of Exercise 1.7 on how to use a table to solve for the variance.

The variance is

$\displaystyle \left(s^2\right)_{aging}=\sum _{i=1}^{10}\:\frac{\left(x_i-\overline{x}\right)^2}{n-1}$

$\displaystyle =\frac{1}{10-1}\left[\left(219-209.90\right)^2+\left(214-209.90\right)^2+...+\left(205-209.90\right)^2\right]$

$\displaystyle =23.62$

The standard deviation is just the square root of the variance. That is

$\displaystyle s_{aging}=\sqrt{s^2}=\sqrt{23.62}=4.86$

Sample Variance & Standard Deviation | Introduction to Statistics and Data Analysis | Probability & Statistics for Engineers & Scientists | Walpole | Problem 1.8

Solution:

We know, based on our answer in Exercise 1.2, that the sample mean is $\displaystyle \overline{x}=20.768$.

To compute for the sample variance, we shall use the formula

$\displaystyle s^2=\sum _{i=1}^n\:\frac{\left(x_i-\overline{x}\right)^2}{n-1}$

The formula states that we need to get the sum of $\displaystyle \left(x_i-\overline{x}\right)^2$, so we can use a table to solve $\displaystyle \left(x_i-\overline{x}\right)^2$ for every sample.

Note: You can refer to the solution of Exercise 1.7 on how to use a table to solve for the variance.

The variance is

$s^2=\frac{1}{20-1}[\left(18.71-20.768\right)^2+\left(21.41-20.768\right)^2$

$\displaystyle +...+\left(21.12-20.768\right)^2]$

$\displaystyle s^2=2.5345$

The standard deviation is just the square root of the variance. That is

$\displaystyle s=\sqrt{s^2}=\sqrt{2.5345}=1.592$

Solution:

We know, based on our answer in Exercise 1.1, that the sample mean is $\displaystyle \overline{x}=3.787$.

To compute for the sample variance, we shall use the formula

$\displaystyle s^2=\sum _{i=1}^n\frac{\left(x_i-\overline{x}\right)^2}{n-1}\:$

The formula states that we need to get the sum of $\displaystyle \left(x_i-\overline{x}\right)^2$, so we can use a table to solve $\displaystyle \left(x_i-\overline{x}\right)^2$ for every sample.

The table above shows that

$\displaystyle \sum _{i=1}^{15}=13.197$

Therefore, the variance is

$\displaystyle s^2=\sum _{i=1}^n\frac{\left(x_i-\overline{x}\right)^2}{n-1}=\frac{13.197}{15-1}=0.9426\:$

The standard deviation is just the square root of the variance. That is

$\displaystyle s=\sqrt{s^2}=\sqrt{0.9426}=0.971$

Mean Computation & Dot Plot | Introduction to Statistics and Data Analysis | Probability & Statistics for Engineers & Scientists | Walpole | Problem 1.6

The tensile strength of silicone rubber is thought to be a function of curing temperature. A study was carried out in which samples of 12 specimens of the rubber were prepared using curing temperatures of 20° C and 45° C. The data below show the tensile strength values in megapascals.

 20° C: 2.07 2.14 2.22 2.03 2.21 2.03 2.05 2.18 2.09 2.14 2.11 2.02 45° C: 2.52 2.15 2.49 2.03 2.37 2.05 1.99 2.42 2.08 2.42 2.29 2.01

Solution:

Part (a)

A dot plot is shown below

In the ﬁgure, “×” represents the 20°C group and “◦” represents the 45°C group.

Part (b)

The mean of the 20°C group is

$\displaystyle \overline{x}_{20^{\circ} C}=2.1075$

The mean of the 45°C group is

$\displaystyle \overline{x}_{45^{\circ} C}=2.2350$

Part (c)

Based on the plot, it seems that high temperature yields more high values of tensile strength, along with a few low values of tensile strength. Overall, the temperature does have an inﬂuence on the tensile strength.

Part (d)

It also seems that the variation of the tensile strength gets larger when the cure temperature is increased.

Introduction to Statistics and Data Analysis | Probability & Statistics for Engineers & Scientists | Walpole | Problem 1.5

Twenty adult males between the ages of 30 and 40 were involved in a study to evaluate the effect of a specific health regimen involving diet and exercise on the blood cholesterol. Ten were randomly selected to be a control group and ten others were assigned to take part in the regimen as the treatment group for a period of 6 months. The following data show the reduction in cholesterol experienced for the time period for the 20 subjects:

 Control Group 7 3 -4 14 2 5 22 -7 9 5 Treatment Group -6 5 9 4 4 12 37 5 3 3

(a) Do a dot plot of the data for both groups on the same graph.

(b) Compute the mean, median, and 10% trimmed means for both groups.

(c) Explain why the difference in the mean suggests one conclusion about the effect of the regimen, while the difference in medians or trimmed means suggests a different conclusion.

Solution:

Part (a)

A dot plot is shown below

Part (b)

The mean, median, and 10% trimmed mean of the control group are

$\displaystyle \overline{x}_{control}=5.60$

$\displaystyle \widetilde{x}_{control}=5.00\:$

$\displaystyle \overline{x}_{tr\left(10\right);control}=5.13$

The mean, median, and 10% trimmed mean of the treatment group are

$\displaystyle \overline{x}_{treatment}=7.60$

$\displaystyle \widetilde{x}_{treatment}=4.50\:$

$\displaystyle \overline{x}_{tr\left(10\right);treatment}=5.63$

Part (c)

The diﬀerence of the means is 2.0 and the diﬀerences of the medians and the trimmed means are 0.5, which are much smaller. The possible cause of this might be due to the extreme values (outliers) in the samples, especially the value of 37.

Introduction to Statistics and Data Analysis | Probability & Statistics for Engineers & Scientists | Walpole | Problem 1.4

In a study conducted by the Department of Mechanical Engineering at Virginia Tech, the steel rods supplied by two different companies were compared. Ten sample springs were made out of the steel rods supplied by each company and a measure of flexibility was recorded for each. The data are as follows:

 Company A: 9.3 8.8 6.8 8.7 8.5 6.7 8 6.5 9.2 7 Company B: 11 9.8 9.9 10.2 10.1 9.7 11 11.1 10.2 9.6

(a) Calculate the sample mean and median for the data for the two companies.

(b) Plot the data for the two companies on the same line and give your impression.

Solution:

Part (a)

The mean and median of Company A are  $\displaystyle \overline{x}_A=7.950\:and\:\widetilde{x}_A=8.250$, respectively.

The mean and median of Company B are  $\displaystyle \overline{x}_B=10.260\:and\:\widetilde{x}_B=10.150$, respectively.

Part (b)

A dot plot is shown below

In the ﬁgure, “×” represents company A and “◦” represents company B. The steel rods made by company B show more ﬂexibility.

Introduction to Statistics and Data Analysis | Probability & Statistics for Engineers & Scientists | Walpole | Problem 1.3

A certain polymer is used for evacuation systems for aircraft. It is important that the polymer be re­sistant to the aging process. Twenty specimens of the polymer were used in an experiment. Ten were assigned randomly to be exposed to the accelerated batch aging process that involved exposure to high tempera­tures for 10 days. Measurements of the tensile strength of the specimens were made and the following data were recorded on tensile strength in psi.

 No aging: 227 222 218 217 225 218 216 229 228 221 Aging: 219 214 215 211 209 218 203 204 201 205

(a) Do a dot plot of the data.

(b) From your plot, does it appear as if the aging pro­cess has had an effect on the tensile strength of this polymer? Explain.

(c) Calculate the sample mean tensile strength of the two samples.

(d) Calculate the median for both. Discuss the simi­larity or lack of similarity between the mean and median of each group.

Introduction to Statistics and Data Analysis | Probability & Statistics for Engineers & Scientists | Walpole | Problem 1.1

The following measurements were recorded for the drying time, in hours, of a certain brand of latex paint

Assume that the measurements are a simple random sample.

a) What is the sample size for the above sample?

b) Calculate the sample mean for this data.

c) Calculate the sample median.

d) Plot the data by way of a dot plot.

e) Compute the 20% trimmed mean for the above data set.