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A random sample of 100 observations from a quantitative population produced a sample mean of 26.8 and a sample standard deviation of \(6.5 .\) Use the \(p\) -value approach to determine whether the population mean is different from \(28 .\) Explain your conclusions.

Short Answer

Expert verified
Answer: No, there is not sufficient evidence to conclude that the population mean is different from 28.

Step by step solution

01

State the null and alternative hypotheses

We begin by stating our null hypothesis (H鈧) and alternative hypothesis (H鈧): H鈧: 渭 = 28 (the population mean is equal to 28) H鈧: 渭 鈮 28 (the population mean is not equal to 28)
02

Calculate the test statistic

We will use a t-statistic for this problem since we are given the sample standard deviation (s) instead of the population standard deviation. The formula for the t-statistic is: t = (x虆 - 渭) / (s /鈭歯) where x虆 is the sample mean, 渭 is the hypothesized population mean, s is the sample standard deviation, and n is the sample size. Plugging in the given values: t = (26.8 - 28) / (6.5 / 鈭100) = (-1.2) / (6.5 / 10) = -1.846
03

Determine the critical value and p-value

To determine the p-value, we need the degrees of freedom, which is equal to the sample size (n) minus 1: df = n - 1 = 100 - 1 = 99 Using a t-distribution table or calculator, we can find the p-value associated with our t-statistic (-1.846) and degrees of freedom (99). Since this is a two-tailed test (H鈧 states 渭 鈮 28), we will find the two-tailed p-value. The p-value in this case is approximately 0.0678.
04

Compare the p-value to the significance level and make a conclusion

Now we need to compare the p-value (0.0678) to a predetermined significance level (伪). The common significance levels are 0.01, 0.05, and 0.10. If the p-value is less than or equal to 伪, we reject the null hypothesis; if the p-value is greater than 伪, we fail to reject the null hypothesis. In this case, let's use a significance level of 0.05. Since the p-value (0.0678) is greater than the significance level (0.05), we fail to reject the null hypothesis. In conclusion, using the p-value approach and a significance level of 0.05, we do not have sufficient evidence to conclude that the population mean is different from 28.

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Key Concepts

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

P-value Approach
Understanding the p-value approach is fundamental in hypothesis testing. It is a method used to provide evidence against the null hypothesis. After conducting a statistical test, the p-value tells us the probability of obtaining a result at least as extreme as the one observed, assuming the null hypothesis is true.

For our problem, we calculate the p-value associated with the observed t-statistic. We then compare this p-value to a predetermined significance level, often denoted by \( \alpha \). Common levels include 0.01, 0.05, or 0.10. If the p-value is less than or equal to the significance level, it suggests that the observed result is unlikely under the null hypothesis, and we may have grounds to reject it. Conversely, a p-value greater than \( \alpha \) means we lack evidence to reject the null hypothesis 鈥 hence, not supporting the alternative hypothesis.
Null and Alternative Hypotheses
Hypothesis testing always begins with the formulation of two competing statements: the null hypothesis \( H_0 \) and the alternative hypothesis \( H_1 \). The null hypothesis is a statement of no effect or no difference and serves as a baseline for the test. In contrast, the alternative hypothesis is what researchers aim to support, indicating some effect or difference.

In our exercise, \( H_0: \mu = 28 \) states that the population mean is equal to 28, while \( H_1: \mu eq 28 \) postulates that the population mean is not 28. Deciding whether to reject \( H_0 \) in favor of \( H_1 \) hinges on the p-value derived from our sample data.
T-statistic
The t-statistic is a ratio that results from dividing the difference between the sample mean and the null hypothesis mean by the standard error of the mean. It's a form of standardization that allows the comparison of the observed data with what we would expect under the null hypothesis.

The formula given is \( t = (\bar{x} - \mu) / (s / \sqrt{n}) \), where \( \bar{x} \) is the sample mean, \( \mu \) is the hypothesized mean, \( s \) is the sample standard deviation, and \( n \) is the sample size. Our calculated t-statistic serves as a measure to ascertain the extremeness of the observed result. The more this value deviates from zero, the more unusual the result is under the null hypothesis.
Sample Mean
The sample mean, often denoted as \( \bar{x} \), is a critical value in hypothesis testing. It represents the average of the observed data points and is essential in calculating the t-statistic. In our exercise, a sample mean of 26.8 is calculated, suggesting where the center of our data lies. This value is used as an estimate of the population mean and is what we compare to the null hypothesis value to gauge whether there are potentially significant differences.
Sample Standard Deviation
The sample standard deviation \( s \) measures the dispersion or variability around the sample mean. It gives us an idea of how spread out the individual measurements are in our sample and is used to calculate the standard error. With our exercise's sample standard deviation being 6.5, it suggests there is a certain amount of variability in the data. A higher standard deviation indicates more variability, influencing the width of our confidence interval and the likelihood of sample means differing from the population mean by chance.

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Most popular questions from this chapter

As a group, students majoring in the engineering disciplines have the highest salary expectations, followed by those studying the computer science fields, according to a Michigan State University study. \({ }^{8}\) To compare the starting salaries of college graduates majoring in electrical engineering and computer science, random samples of 50 recent college graduates in each major were selected and the following information was obtained: $$ \begin{array}{lcc} \hline \text { Major } & \text { Mean (\$) } & \text { SD } \\ \hline \text { Electrical Engineering } & 62,428 & 12,500 \\ \text { Computer Science } & 57,762 & 13,330 \\ \hline \end{array} $$ a. Do the data provide sufficient evidence to indicate a difference in average starting salaries for college graduates who majored in electrical engineering and computer science? Test using \(\alpha=.05\). b. Calculate a \(95 \%\) confidence interval for the difference in the two population means. Does this confirm your conclusion in part a? Explain.

Early Detection of Breast Cancer Of those women who are diagnosed to have early-stage breast cancer, one-third eventually die of the disease. Suppose a screening program for the early detection of breast cancer was started in order to increase the survival rate \(p\) of those diagnosed to have the disease. A random sample of 200 women was selected from among those who were screened by the program and who were diagnosed to have the disease. Let \(x\) represent the number of those in the sample who survive the disease. a. If you wish to determine whether the screening program has been effective, state the alternative hypothesis that should be tested. b. State the null hypothesis. c. If 164 women in the sample of 200 survive the disease, can you conclude that the screening program was effective? Test using \(\alpha=.05\) and explain the practical conclusions from your test. d. Find the \(p\) -value for the test and interpret it.

Calculate the p-value for the hypothesis tests given. $$ n=1000 \text { and } x=279 \text { . You wish to show that } p<.3 . $$

Independent random samples were selected from two binomial populations, with sample sizes and the number of successes given. Use this information to calculate \(\hat{p}_{1}, \hat{p}_{2},\) and \(\hat{p}\). $$ n_{1}=60, n_{2}=60, x_{1}=43, x_{2}=36 $$

It is reported \(^{l}\) that the average or mean number of Facebook friends is \(155 .\) Suppose that when 50 randomly chosen Facebook users are polled regarding the number of their friends, the average number of their friends was reported to be \(\bar{x}=149\) with a standard deviation of \(s=29.7\). Use this information to answer the questions in Exercises \(13-15 .\) Calculate the value of the \(z\) -statistic based on the sample mean, \(\bar{x}\). Is this an unusual value of \(z\) ?

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