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Consider the following hypothesis test: \\[ \begin{array}{l} H_{0}: \mu \leq 12 \\ H_{\mathrm{a}}: \mu>12 \end{array} \\] A sample of 25 provided a sample mean \(\bar{x}=14\) and a sample standard deviation \(s=4.32\) a. Compute the value of the test statistic. b. Use the \(t\) distribution table (Table 2 in Appendix \(B\) ) to compute a range for the \(p\) -value. c. \(\quad\) At \(a=.05,\) what is your conclusion? d. What is the rejection rule using the critical value? What is your conclusion?

Short Answer

Expert verified
The hypothesis test concludes that \( \mu > 12 \).

Step by step solution

01

Identify the parameters and the test statistic formula

We are given that \( n = 25 \), the sample mean \( \bar{x} = 14 \), and the sample standard deviation \( s = 4.32 \). The null hypothesis \( H_0 \) is that \( \mu \leq 12 \) and the alternative hypothesis \( H_a \) is that \( \mu > 12 \). The test statistic for a one-sample \( t \)-test is given by:\[t = \frac{\bar{x} - \mu_0}{s/\sqrt{n}}\]Where \( \mu_0 \) is the hypothesized population mean, which is 12 in this case.
02

Calculate the test statistic

Substitute the known values into the test statistic formula:\[t = \frac{14 - 12}{4.32/\sqrt{25}} = \frac{2}{0.864} \approx 2.31\]Thus, the test statistic \( t \approx 2.31 \).
03

Find the degrees of freedom

For a sample size of \( n = 25 \), the degrees of freedom (\( df \)) is given by \( df = n - 1 = 24 \).
04

Determine the p-value range

Using a t distribution table for \( df = 24 \), locate the test statistic value \( t = 2.31 \). It falls between the table's \( t \) values for \( p = 0.025 \) and \( p = 0.01 \). Thus, the p-value for this test statistic is in the range \( 0.01 < p < 0.025 \).
05

Compare p-value to significance level

The significance level \( \alpha = 0.05 \). Since the p-value range \( 0.01 < p < 0.025 \) is lower than \( 0.05 \), we reject the null hypothesis \( H_0 \).
06

Determine rejection rule using the critical value

The critical value for a one-tailed test with \( \alpha = 0.05 \) and \( df = 24 \) is approximately \( t_{\alpha} \approx 1.711 \). Since the calculated \( t \)-value (2.31) is greater than 1.711, we reject the null hypothesis.
07

State the final conclusion

Both the p-value method and critical value method result in rejecting the null hypothesis. Therefore, there is sufficient evidence to conclude that \( \mu > 12 \).

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Key Concepts

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

t-distribution
The t-distribution is a probability distribution that is used when estimating population parameters when the sample size is small and the population standard deviation is unknown. It is particularly useful in hypothesis testing for small sample sizes, typically less than 30. In the context of this problem, since we have a sample size of 25, the t-distribution is appropriate.
The shape of the t-distribution is similar to the normal distribution but has thicker tails, meaning it gives larger probabilities to extreme values. This is especially important in hypothesis testing because it accounts for the extra variability introduced when estimating a population standard deviation from a sample.
  • The test statistic in a t-distribution is calculated as: \[t = \frac{\bar{x} - \mu_0}{s/\sqrt{n}}\]where \( \bar{x} \) is the sample mean, \( \mu_0 \) is the null hypothesis mean, \( s \) is the sample standard deviation, and \( n \) is the sample size.
  • Degrees of freedom \( df = n - 1 \) are crucial for the t-distribution, defining its shape. For instance, a larger \( df \) will make the t-distribution more closely resemble a normal distribution.
p-value
The p-value is a critical concept in hypothesis testing. It measures the probability of observing test results at least as extreme as the results actually observed, under the assumption that the null hypothesis is true. In simpler terms, it's a way to quantify evidence against a null hypothesis.
In the exercise example, with the null hypothesis \( H_0: \mu \leq 12 \), the calculated t-test statistic was approximately 2.31. Using this value, we determined its position on the t-distribution with \( df = 24 \) to find a p-value range.
  • A small p-value (typically \(< 0.05\) ) indicates strong evidence against the null hypothesis, so we reject \( H_0 \).
  • The calculated p-value in this example was between 0.01 and 0.025, which is less than the commonly used significance level of 0.05, leading us to reject \( H_0 \).
  • A p-value does not tell us the probability that the null hypothesis is true or false. Instead, it indicates how statistically significant the test results are.
significance level
The significance level in hypothesis testing, often denoted as \( \alpha \), is the threshold used to determine whether a hypothesis test's result is statistically significant. It represents the probability of rejecting the null hypothesis if it is in fact true, also known as a Type I error.
Common choices for \( \alpha \) include 0.05, 0.01, or 0.10. In this exercise, a significance level of 0.05 was chosen. This means there is a 5% risk of incorrectly rejecting the null hypothesis.
  • If the p-value found from the test statistic is less than or equal to the significance level, we reject the null hypothesis. This suggests that there is strong enough evidence to support the alternative hypothesis.
  • If the p-value is higher than \( \alpha \), we fail to reject the null hypothesis.
Understanding the role of the significance level helps in making decisions based on statistical tests. In the exercise, because our p-value range was less than 0.05, we concluded evidence supports that \( \mu > 12\).

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