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Use the following information to answer the next five exercises. A study was conducted to test the effectiveness of a software patch in reducing system failures over a six-month period. Results for randomly selected installations are shown in Table 10.21. The 鈥渂efore鈥 value is matched to an 鈥渁fter鈥 value, and the differences are calculated. The differences have a normal distribution. Test at the 1% significance level. $$ \begin{array}{|l|l|l|l|l|l|l|}\hline \text { Installation } & {\mathbf{A}} & {\mathbf{B}} & {\mathbf{C}} & {\mathbf{D}} & {\mathbf{E}} & {\mathbf{F}} & {\mathbf{G}} & {\mathbf{H}} \\ \hline \text { Before } & {3} & {6} & {4} & {2} & {5} & {8} & {2} & {6} \\ \hline \text { After } & {1} & {5} & {2} & {0} & {1} & {0} & {2} & {2} \\ \hline\end{array} $$ What is the random variable?

Short Answer

Expert verified
The random variable is the difference in system failures before and after the patch for each installation.

Step by step solution

01

Understand the Context

A study is conducted to test the effectiveness of a software patch. It measures system failures before and after applying the software patch to several installations. Results show the number of failures before and after the patch.
02

Identify Pairwise Data

For each installation (A through H), the number of failures is recorded before and after the patch. This creates pairs of data points for the 'before' and 'after' conditions for each installation.
03

Define the Difference

The difference for each installation is calculated as the number of failures 'before' minus the number of failures 'after'. This difference measures the change in system failures due to the patch.
04

Identify the Random Variable

The random variable is typically the measurement or outcome being analyzed for variation or change. Here, it is the 'difference in number of system failures between before and after applying the patch for each installation.' These differences are assumed to follow a normal distribution for the test.

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Key Concepts

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

Paired Sample
When dealing with data that comes in pairs, especially when analyzing the effectiveness of treatments or conditions, we use a paired sample approach. This method is particularly useful because it can effectively control for variability between subjects that is not related to the treatment or condition.
In our exercise, each installation is treated as a subject, and we collect data in pairs: once before the software patch is applied, and once after. By comparing these paired measurements, we can directly assess the impact of the software patch on each installation without the interference of other variables.
  • Poor performance before the patch vs after in multiple installations
  • Directly correlating system failures in tangible way
  • More reliable conclusions by focusing on differences within the same subject
Normal Distribution
Understanding the normal distribution is crucial when performing hypothesis testing, particularly with paired samples. A normal distribution is a bell-shaped curve that is symmetrical about the mean. Most data points lie close to the mean, with fewer points appearing as you move further away from it. In the context of our study, we assume that the differences in system failures before and after applying the software patch follow a normal distribution. This assumption allows us to use statistical methods that require normally distributed data to draw conclusions about the treatment's effectiveness. However, keep in mind:
  • Sufficient sample size helps ensure normal distribution
  • Normalized data relies on the assumption that external factors minimally affect results
  • Helps in utilizing mathematical tools like the t-test for analysis
Significance Level
Significance levels help us decide how much risk of being wrong we are willing to take in our hypothesis testing. In simpler terms, it tells us how confident we can be in the results. Commonly used significance levels are 1%, 5%, or 10%. For our study, the significance level is set at 1%, meaning there is a 1% chance that our results are due to random variation rather than the software patch making a true difference. This is a stringent level, ensuring that our conclusions are very robust.
  • 1% significance level equates to higher confidence in the patch's effectiveness
  • Helps in demonstrating whether observed effect is meaningful
  • Balances the risk of making Type I errors (false positives)

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Most popular questions from this chapter

Use the following information for the next five exercises. Two types of phone operating system are being tested to determine if there is a difference in the proportions of system failures (crashes). Fifteen out of a random sample of 150 phones with OS1 had system failures within the first eight hours of operation. Nine out of another random sample of 150 phones with OS2 had system failures within the first eight hours of operation. OS2 is believed to be more stable (have fewer crashes) than OS1. State the null and alternative hypotheses.

Use the following information to answer next five exercises. A study was conducted to test the effectiveness of a juggling class. Before the class started, six subjects juggled as many balls as they could at once. After the class, the same six subjects juggled as many balls as they could. The differences in the number of balls are calculated. The differences have a normal distribution. Test at the 1% significance level. $$ \begin{array}{|c|c|c|c|c|c|}\hline \text { Subject } & {\mathbf{A}} & {\mathbf{B}} & {\mathbf{c}} & {\mathbf{D}} & {\mathbf{E}} & {\mathbf{F}} \\\ \hline \text { Before } & {3} & {4} & {3} & {2} & {4} & {5} \\ \hline \text { After } & {4} & {5} & {6} & {4} & {5} & {7} \\ \hline\end{array} $$ What is the p-value?

Use the following information to answer the next five exercises. A doctor wants to know if a blood pressure medication is effective. Six subjects have their blood pressures recorded. After twelve weeks on the medication, the same six subjects have their blood pressure recorded again. For this test, only systolic pressure is of concern. Test at the 1% significance level. $$ \begin{array}{|l|l|l|l|l|l|}\hline \text { Patient } & {\mathbf{A}} & {\mathbf{B}} & {\mathbf{C}} & {\mathbf{D}} & {\mathbf{E}} & {\mathbf{F}} \\\ \hline \text { Before } & {161} & {162} & {165} & {162} & {166} & {171} \\\ \hline \text { After } & {158} & {159} & {166} & {160} & {167} & {169} \\\ \hline\end{array} $$ What is the test statistic?

Use the following information to answer the next 12 exercises: The U.S. Center for Disease Control reports that the mean life expectancy was 47.6 years for whites born in 1900 and 33.0 years for nonwhites. Suppose that you randomly survey death records for people born in 1900 in a certain county. Of the 124 whites, the mean life span was 45.3 years with a standard deviation of 12.7 years. Of the 82 nonwhites, the mean life span was 34.1 years with a standard deviation of 15.6 years. Conduct a hypothesis test to see if the mean life spans in the county were the same for whites and nonwhites. At a pre-conceived \(\alpha=0.05,\) what is your: a. Decision: b. Reason for the decision: c. Conclusion (write out in a complete sentence):

Parents of teenage boys often complain that auto insurance costs more, on average, for teenage boys than for teenage girls. A group of concerned parents examines a random sample of insurance bills. The mean annual cost for 36 teenage boys was \(679. For 23 teenage girls, it was \)559. From past years, it is known that the population standard deviation for each group is $180. Determine whether or not you believe that the mean cost for auto insurance for teenage boys is greater than that for teenage girls.

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