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Chapter 11: Problem 8

For each of the two samples $$ \begin{array}{l} \text { Sarmla } 1:\\{0.8,1.2,1.1,0.8,-2.0\\} \\ \text { Sarmle } 2:\\{-2.0,0.0,0.7,0.8,2.2,4.1,-1.8\\} \end{array} $$ find a. the sample size, b. the sample mean, c. the sample variance.

Short Answer

Expert verified
Sample 1: Size = 5, Mean = 0.38, Variance = 1.795; Sample 2: Size = 7, Mean 鈮� 0.57, Variance 鈮� 5.02.

Step by step solution

01

Determine the Sample Size for Sample 1

The sample size is simply the number of observations in the sample. For Sample 1, count the numbers provided: \(0.8, 1.2, 1.1, 0.8, -2.0\). The total number of values is 5, so the sample size is 5.
02

Determine the Sample Size for Sample 2

Similarly, for Sample 2, count the numbers provided: \(-2.0, 0.0, 0.7, 0.8, 2.2, 4.1, -1.8\). The total number of values is 7, so the sample size is 7.
03

Calculate the Sample Mean for Sample 1

The sample mean is calculated by adding all the values and dividing by the sample size. For Sample 1: \( mean = \frac{0.8 + 1.2 + 1.1 + 0.8 - 2.0}{5} = \frac{1.9}{5} = 0.38 \).
04

Calculate the Sample Mean for Sample 2

For Sample 2: Calculate the mean by adding all values and dividing by the sample size: \( mean = \frac{-2.0 + 0.0 + 0.7 + 0.8 + 2.2 + 4.1 - 1.8}{7} = \frac{4.0}{7} \approx 0.57 \).
05

Calculate the Sample Variance for Sample 1

Sample variance is calculated as follows: \( variance = \frac{\sum_{i=1}^{n} (x_i - \overline{x})^2}{n-1} \). For Sample 1, \( x_i - \overline{x} \) yields \((0.8-0.38)^2, (1.2-0.38)^2, (1.1-0.38)^2, (0.8-0.38)^2, (-2.0-0.38)^2\), and the variance is \( \frac{(0.42)^2 + (0.82)^2 + (0.72)^2 + (0.42)^2 + (-2.38)^2}{4} = \frac{7.18}{4} = 1.795 \).
06

Calculate the Sample Variance for Sample 2

Follow similar steps to calculate for Sample 2: \( x_i - \overline{x} \) values become \((-2.0 - 0.57)^2, (0.0 - 0.57)^2, (0.7 - 0.57)^2, (0.8 - 0.57)^2, (2.2 - 0.57)^2, (4.1 - 0.57)^2, (-1.8 - 0.57)^2\). The variance is \( \frac{(2.57)^2 + (0.57)^2 + (0.13)^2 + (0.23)^2 + (1.63)^2 + (3.53)^2 + (2.37)^2}{6} = \frac{30.13}{6} \approx 5.02 \).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Sample Size
The sample size is a fundamental aspect of descriptive statistics. It simply refers to the number of observations or data points contained within a sample. Determining the sample size is often the first step when analyzing data. This straightforward step involves counting the elements in your data set.

Consider Sample 1 with the following data: \(0.8, 1.2, 1.1, 0.8, -2.0\). To find the sample size, count these numbers. Here, you have 5 observations, so the sample size is 5.

For Sample 2, the data is \(-2.0, 0.0, 0.7, 0.8, 2.2, 4.1, -1.8\). Count these, and you find there are 7 numbers, making the sample size 7.
  • Sample Size for Sample 1: 5
  • Sample Size for Sample 2: 7
Remember, the sample size is an easy but critical starting point in statistical analysis.
Sample Mean
The sample mean offers a way to summarize a set of data using a single value, representing the average of the numbers in a sample. This is calculated by summing all the data points and dividing by the sample size. The sample mean provides valuable insights into the central tendency of data.

For the data in Sample 1, \( mean = \frac{0.8 + 1.2 + 1.1 + 0.8 - 2.0}{5} = 0.38 \). The calculation shows you add the values together to get a total sum, then divide by 5 (the sample size) to find the sample mean, which is 0.38.

In Sample 2, the means are calculated similarly: \( mean = \frac{-2.0 + 0.0 + 0.7 + 0.8 + 2.2 + 4.1 - 1.8}{7} \approx 0.57 \). The corresponding mean here is approximately 0.57.
  • Sample Mean for Sample 1: 0.38
  • Sample Mean for Sample 2: 0.57
The sample mean is crucial for understanding your data's overall level or typical value.
Sample Variance
Sample variance measures how much the data points in a sample spread out from the average (or mean) value. It is useful in assessing the consistency of data. The larger the variance, the more dispersed the data. To calculate it, find the difference between each data point and the mean, square these differences, sum them, and finally divide by one less than the sample size.

For Sample 1: Calculate the differences \((x_i - \overline{x})^2\) as \((0.8-0.38)^2, (1.2-0.38)^2,...\) and find \( variance = \frac{7.18}{4} = 1.795 \).

Similarly, for Sample 2, apply the formula. Calculate \((x_i - \overline{x})^2\) for each observation, add these squared values, and you find \( variance = \frac{30.13}{6} \approx 5.02 \).
  • Sample Variance for Sample 1: 1.795
  • Sample Variance for Sample 2: 5.02
Understanding sample variance helps in determining the variability within your dataset, giving you deeper insights into your data's behavior.

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Most popular questions from this chapter

A laptop computer maker uses battery packs supplied by two companies, Aand B. While both brands have the same average battery life between charges (LBC), the computer maker seems to receive more complaints about shorter \(\mathrm{LBC}\) than expected for battery packs supplied by company \(\mathrm{B}\). The computer maker suspects that this could be caused by higher variance in \(\mathrm{LBC}\) for Brand \(B\). To check that, ten new battery packs from each brand are selected, installed on the same models of laptops, and the laptops are allowed to run until the battery packs are completely discharged. The following are the observed LBCs in hours. Test, at the \(5 \%\) level of significance, whether the data provide sufficient evidence to conclude that the LBCs of Brand \(B\) have a larger variance that those of Brand \(A\).

The Department of Parks and Wildlife stocks a large lake with fish every six years. It is determined that a healthy diversity of fish in the lake should consist of \(10 \%\) largemouth bass, \(15 \%\) smallmouth bass, \(10 \%\) striped bass, \(10 \%\) trout, and \(20 \%\) catfish. Therefore each time the lake is stocked, the fish population in the lake is restored to maintain that particular distribution. Every three years, the department conducts a study to see whether the distribution of the fish in the lake has shifted away from the target proportions. In one particular year, a research group from the department observed a sample of 292 fish from the lake with the results given in the table provided. Test, at the \(5 \%\) level of significance, whether there is sufficient evidence in the data to conclude that the fish population distribution has shifted since the last stocking. $$\begin{array}{|c|c|c|} \hline \text { Fish } & \text { Target Distribution } & \text { Fish in Sample } \\ \hline \text { Largemouth Bass } & 0.10 & 14 \\ \hline \text { Smallmouth Bass } & 0.15 & 49 \\ \hline \text { Striped Bass } & 0.10 & 21 \\ \hline \text { Trout } & 0.10 & 22 \\ \hline \text { Catfish } & 0.20 & 75 \\ \hline \text { Other } & 0.35 & 111\\\ \hline \end{array}$$

For each of the two samples $$ \begin{array}{l} \text { Sample } 1:\\{8,1,11,0,-2,\\} \\ \text { Sarrpla } 2:\\{-2,0,0,0,2,4,-1\\} \end{array} $$ find a. the sample size, b. the sample mean, c. the sample variance.

The Mozart effect refers to a boost of average performance on tests for elementary school students if the students listen to Mozart's chamber music for a period of time immediately before the test. Many educators believe that such an effect is not necessarily due to Mozart's music per se but rather a relaxation period before the test. To support this belief, an elementary school teacher conducted an experiment by dividing her third-grade class of 15 students into three groups of \(5 .\) Students in the first group were asked to give themselves a self-administered facial massage; students in the second group listened to Mozart's chamber music for 15 minutes; students in the third group listened to Schubert's chamber music for 15 minutes before the test. The scores of the 15 students are given below: $$ \begin{array}{|c|c|c|} \hline \text { Group 1 } & \text { Group 2 } & \text { Group 3 } \\ \hline 79 & 82 & 80 \\ \hline 81 & 84 & 81 \\ \hline 80 & 86 & 71 \\ \hline 89 & 91 & 90 \\ \hline 86 & 82 & 86 \\ \hline \end{array} $$ Test, using the ANOVA F-test at the \(10 \%\) level of significance, whether the data provide sufficient evidence to conclude that any of the three relaxation method does better than the others.

Japanese sturgeon is a subspecies of the sturgeon family indigenous to Japan and the Northwest Pacific. In a particular fish hatchery newly hatched baby Japanese sturgeon are kept in tanks for several weeks before being transferred to larger ponds. Dissolved oxygen in tank water is very tightly monitored by an electronic system and rigorously maintained at a target level of 6.5 milligrams per liter (mg/I). The fish hatchery looks to upgrade their water monitoring systems for tighter control of dissolved oxygen. A new system is evaluated against the old one currently being used in terms of the variance in measured dissolved oxygen. Thirty-one water samples from a tank operated with the new system were collected and 16 water samples from a tank operated with the old system were collected, all during the course of a day. The samples yield the following information: New Sample \(1: n 1=31 s 21=0.0121\) $$ \text { Old Sample } 2: n_{2}=16 \mathrm{~s} 22=0.0319 $$ Test, at the \(10 \%\) level of significance, whether the data provide sufficient evidence to conclude that the new system will provide a tighter control of dissolved oxygen in the tanks.
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