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Chapter 9: Problem 16

A random sample of \(n=35\) observations from a quantitative population produced a mean \(\bar{x}=2.4\) and a standard deviation of \(s=.29 .\) Your research objective is to show that the population mean \(\mu\) exceeds 2.3. Use this information to answer the questions. Do the data provide sufficient evidence to conclude that \(\mu>2.3 ?\)

Short Answer

Expert verified
Answer: Yes, the data provides sufficient evidence to conclude that the population mean is greater than 2.3, as the t-test statistic (2.1065) is greater than the critical value (1.690) at a 0.05 significance level.

Step by step solution

01

Define the hypotheses

We define our two hypotheses as follows: Null hypothesis (\(H_0\)): \(\mu = 2.3\) Alternative hypothesis (\(H_1\)): \(\mu > 2.3\)
02

Calculate the t-test statistic

The t-test statistic can be calculated using the formula: \(t = \frac{\bar{x} - \mu_0}{\frac{s}{\sqrt{n}}}\), where \(\bar{x}\) is the sample mean, \(\mu_0\) is the hypothesized population mean, \(s\) is the sample standard deviation, and \(n\) is the sample size. In this case, we have \(\bar{x} = 2.4\), \(\mu_0 = 2.3\), \(s = 0.29\), and \(n = 35\). Plugging in these values, we get: \(t = \frac{2.4 - 2.3}{\frac{0.29}{\sqrt{35}}} = 2.1065\)
03

Determine the critical value and compare it to the t-test statistic

To find the critical value, we need to determine the degrees of freedom (\(df\)), which is equal to \(n - 1\) and then choose a significance level, which is usually denoted as alpha (\(\alpha\)). Common values for \(\alpha\) are 0.05 and 0.01. Here, we have \(df = 35 - 1 = 34\). Let's use a significance level of \(\alpha = 0.05\). Now, we can find the critical value from the t-distribution table or using a calculator/software. For a one-tailed test with \(df = 34\) and \(\alpha = 0.05\), the critical value is approximately 1.690. Now, we compare the t-test statistic (2.1065) to the critical value (1.690). Since \(t >\) critical value, we reject the null hypothesis.
04

Conclusion

Based on the results of the t-test, with a t-test statistic of 2.1065 and a critical value of 1.690, we reject the null hypothesis (\(H_0: \mu = 2.3\)) in favor of the alternative hypothesis (\(H_1: \mu > 2.3\)). Therefore, the data provides sufficient evidence to conclude that the population mean \(\mu > 2.3\).

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Key Concepts

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

Null Hypothesis
Understanding the null hypothesis is crucial when performing a t-test. It represents the default claim that there is no effect or no difference. In statistical terms, the null hypothesis (\(H_0\)) asserts that any observed effect is due to chance rather than a genuine relationship. For example, if we are testing whether a new tutoring program improves student test scores, the null hypothesis would be that the program has no effect on scores.
Alternative Hypothesis
In contrast to the null hypothesis, the alternative hypothesis (\(H_1\) or \(H_a\)) posits that there is an effect or a difference that is not due to random chance. It's what the researcher aims to support. Continuing our earlier example, the alternative hypothesis would suggest that the tutoring program does have a measurable positive impact on student test scores. This hypothesis is considered only if the null hypothesis is rejected after statistical testing.
Critical Value
The critical value is a key concept in hypothesis testing. It is the threshold against which the test statistic is compared to help us decide whether to reject the null hypothesis. The critical value depends on the chosen significance level (\( \text{alpha} \(\alpha\)\) and the degrees of freedom in the test. Typically, a significance level of 0.05 is used, indicating a 5% risk of concluding that an effect exists when it does not. If the test statistic exceeds the critical value, the null hypothesis is rejected, pointing towards an effect or difference that is statistically significant.

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Most popular questions from this chapter

A random sample of \(n=35\) observations from a quantitative population produced a mean \(\bar{x}=2.4\) and a standard deviation of \(s=.29 .\) Your research objective is to show that the population mean \(\mu\) exceeds 2.3. Use this information to answer the questions. Repeat the calculation of \(\beta\) for \(\mu=2.3,2.5,\) and \(2.6 .\)

Independent random samples of 280 and 350 observations were selected from binomial populations 1 and 2, respectively. Sample 1 had 132 successes, and sample 2 had 178 successes. Do the data present sufficient evidence to indicate that the proportion of successes in population 1 is smaller than the proportion in population 2 ? Use one of the two methods of testing presented in this section, and explain your conclusions.

For the situations described in Exercises \(7-10,\) state the null and alternative hypotheses to be tested. A seed wash is expected to change the proportion \(p\) of seeds that germinate when planted in poorlydraining soil. Suppose that the current germination rate is about \(80 \%\).

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 \(\hat{p}\). $$ n_{1}=60, n_{2}=60, x_{1}=43, x_{2}=36 $$

Independent random samples of 140 observations were randomly selected from binomial populations 1 and 2 , respectively. Sample 1 had 74 successes and sample 2 had 81 successes. Use this information to answer the questions. Suppose that, for practical reasons, you know that \(p\), cannot be larger than \(p_{2}\). Test the appropriate hypothesis using \(\alpha=.10\).
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