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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 p-value?

Short Answer

Expert verified
p-value is calculated from t-statistic with 7 degrees of freedom using a two-tailed test.

Step by step solution

01

Calculate Difference

Calculate the difference between the 'before' and 'after' values for each installation. The differences are: \( A: 3-1=2 \), \( B: 6-5=1 \), \( C: 4-2=2 \), \( D: 2-0=2 \), \( E: 5-1=4 \), \( F: 8-0=8 \), \( G: 2-2=0 \), \( H: 6-2=4 \).
02

Compute Mean and Standard Deviation

Find the mean and standard deviation of the differences. Mean \( \bar{x} = \frac{2+1+2+2+4+8+0+4}{8} = 2.875 \). Standard deviation \( s \) is computed using the formula for standard deviation for sample data.
03

Set Null and Alternative Hypothesis

The null hypothesis \( H_0 \) is that the mean difference is zero, \( \mu_d = 0 \), and the alternative hypothesis \( H_a \) is that the mean difference is not zero, \( \mu_d eq 0 \).
04

Calculate t-Statistic

Use the formula \( t = \frac{\bar{x} - \mu_0}{s/\sqrt{n}} \), where \( \mu_0 = 0 \), \( \bar{x} = 2.875 \), and \( n = 8 \). Plug in the values to find the t-statistic.
05

Determine Degrees of Freedom

Degrees of freedom \( df = n - 1 = 8 - 1 = 7 \).
06

Find the p-value

Use the t-distribution table, calculator, or statistical software to find the p-value for the calculated t-statistic with 7 degrees of freedom. Look for the two-tailed p-value since the test is non-directional.

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Key Concepts

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

p-value calculation
The p-value is a crucial concept in hypothesis testing that helps determine whether the observed data fits the assumption described in the null hypothesis. In simple terms, a p-value tells us how likely it is to get a result as extreme as the sample result, assuming the null hypothesis is true.

To calculate the p-value in this particular exercise, we need to use the t-statistic calculated from the sample data. Once we have the t-value, we look it up in a t-distribution table or use statistical software to find the p-value, which corresponds to the probability of observing a test statistic at least as extreme as the one calculated.

It's important to note that the p-value is directly related to the test's significance level. If the p-value is less than the significance level, we reject the null hypothesis. In this exercise, the task is to determine the p-value at the 1% significance level, to assess whether there is a significant reduction in system failures after applying the software patch.
t-statistic
The t-statistic is a type of statistic used to determine if there is a significant difference between the means of two related groups. In our case, it helps to identify if there鈥檚 a difference in system failures before and after applying a patch.

To calculate the t-statistic, first, we need the mean of the differences between the 'before' and 'after' values. Then, we use the formula:\[t = \frac{\bar{x} - \mu_0}{s/\sqrt{n}}\]Where \( \bar{x} \) is the mean of the differences, \( \mu_0 \) is the hypothesized population mean difference (typically zero under the null hypothesis), \( s \) is the standard deviation of the differences, and \( n \) is the sample size.

In this context, the t-statistic quantifies how much the mean difference is away from zero (the null hypothesis), considering the variability of the data and the sample size. A higher magnitude of the t-statistic indicates a larger difference that is less likely due to random variation.
paired sample t-test
The paired sample t-test is a statistical method used to compare two related samples. These samples can be 'before' and 'after' measures taken from the same group, like in this exercise where we're observing system failures before and after applying a software patch.

The test helps to determine if the mean of the differences between the paired observations is significantly different from zero. It is called 'paired' because the observations are not independent; each 'before' score is directly paired with an 'after' score.

The major advantages of the paired sample t-test are that it controls for within-subject variability and reduces the impact of extraneous variables because each subject serves as their own control. This makes the test more sensitive at detecting differences than an unpaired test.
significance level
The significance level, often denoted by \( \alpha \), is a threshold chosen by the researcher to determine how confident they want to be in rejecting the null hypothesis. It quantifies the probability of committing a Type I error, which is the mistake of rejecting a true null hypothesis.

In practice, common choices for \( \alpha \) include 0.05, 0.01, and 0.1, with smaller values indicating stricter criteria for rejecting the null hypothesis. In this problem, a significance level of 0.01 means that there is only a 1% chance of rejecting the null hypothesis if it is indeed true.

Selecting a significance level involves balancing the risks of making errors. A low significance level means you require strong evidence to reject the null hypothesis, thus reducing the risk of a false positive. However, it also increases the likelihood of failing to detect a true effect, known as Type II error. This trade-off should be considered based on the specific context and consequences of incorrect decisions in the study.

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Most popular questions from this chapter

Use the following information to answer the next ten exercises. indicate which of the following choices best identifies the hypothesis test. a. independent group means, population standard deviations and/or variances known b. independent group means, population standard deviations and/or variances unknown c. matched or paired samples d. single mean e. two proportions f. single proportion A new chocolate bar is taste-tested on consumers. Of interest is whether the proportion of children who like the new chocolate bar is greater than the proportion of adults who like it.

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 the next ten exercises. indicate which of the following choices best identifies the hypothesis test. a. independent group means, population standard deviations and/or variances known b. independent group means, population standard deviations and/or variances unknown c. matched or paired samples d. single mean e. two proportions f. single proportion A football league reported that the mean number of touchdowns per game was five. A study is done to determine if the mean number of touchdowns has decreased.

Use the following information to answer the next three exercises. Neuroinvasive West Nile virus is a severe disease that affects a person鈥檚 nervous system . It is spread by the Culex species of mosquito. In the United States in 2010 there were 629 reported cases of neuroinvasive West Nile virus out of a total of 1,021 reported cases and there were 486 neuroinvasive reported cases out of a total of 712 cases reported in 2011. Is the 2011 proportion of neuroinvasive West Nile virus cases more than the 2010 proportion of neuroinvasive West Nile virus cases? Using a 1% level of significance, conduct an appropriate hypothesis test. 鈥 鈥2011鈥 subscript: 2011 group. 鈥 鈥2010鈥 subscript: 2010 group The p-value is 0.0022. At a 1% level of significance, the appropriate conclusion is a. There is sufficient evidence to conclude that the proportion of people in the United States in 2011 who contracted neuroinvasive West Nile disease is less than the proportion of people in the United States in 2010 who contracted neuroinvasive West Nile disease. b. There is insufficient evidence to conclude that the proportion of people in the United States in 2011 who contracted neuroinvasive West Nile disease is more than the proportion of people in the United States in 2010 who contracted neuroinvasive West Nile disease. c. There is insufficient evidence to conclude that the proportion of people in the United States in 2011 who contracted neuroinvasive West Nile disease is less than the proportion of people in the United States in 2010 who contracted neuroinvasive West Nile disease. d. There is sufficient evidence to conclude that the proportion of people in the United States in 2011 who contracted neuroinvasive West Nile disease is more than the proportion of people in the United States in 2010 who contracted neuroinvasive West Nile disease

Joan Nguyen recently claimed that the proportion of college-age males with at least one pierced ear is as high as the proportion of college-age females. She conducted a survey in her classes. Out of 107 males, 20 had at least one pierced ear. Out of 92 females, 47 had at least one pierced ear. Do you believe t at the proportion of males has reached the proportion of females?

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