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Given the following hypothesis: $$\begin{array}{l}H_{0}: \mu \leq 10 \\\H_{1}: \mu>10\end{array}$$ For a random sample of 10 observations, the sample mean was 12 and the sample standard deviation \(3 .\) Using the .05 significance level: a. State the decision rule. b. Compute the value of the test statistic. c. What is your decision regarding the null hypothesis?

Short Answer

Expert verified
Reject \(H_0\); the test statistic \(t = 2.11\) is greater than the critical value 1.833.

Step by step solution

01

Determine the decision rule

We are testing a single mean, so we will use a t-test. Since the sample size is less than 30, the t-distribution applies. For the hypothesis test, \[ H_{0}: \mu \leq 10 \quad \text{and} \quad H_{1}: \mu >10 \]we use a right-tailed test. Given the significance level \(\alpha = 0.05\) and degrees of freedom \(df = n-1 = 9\), we find the critical t-value, \(t_{\alpha} = t_{0.05,9}\), using a t-distribution table or calculator. This value is approximately 1.833. The decision rule will be to reject \(H_{0}\) if the calculated test statistic is greater than 1.833.
02

Calculate the test statistic

The formula for the t-test statistic is \[ t = \frac{\bar{x} - \mu}{s/\sqrt{n}} \]where \(\bar{x}\) is the sample mean, \(\mu\) is the population mean under \(H_{0}\), \(s\) is the sample standard deviation, and \(n\) is the sample size. Plug in the values:\[ t = \frac{12 - 10}{3/\sqrt{10}} = \frac{2}{0.9487} \approx 2.11 \]
03

Make a decision regarding the null hypothesis

Compare the calculated test statistic to the critical t-value from Step 1. Since 2.11 > 1.833, we reject the null hypothesis \(H_{0}\). This suggests there is sufficient evidence at the 0.05 significance level to conclude that the mean is greater than 10.

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Key Concepts

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

t-test
The t-test is a statistical method used to determine if there is a significant difference between the mean of a sample and a known value or the mean from another sample. It is especially useful when dealing with small sample sizes (typically less than 30). In our exercise, we conducted a one-sample t-test because we aimed to compare the sample mean to a specific population mean, which under our null hypothesis is 10.
When using a t-test, especially with small samples, we rely on the t-distribution rather than the normal distribution. This helps to account for the increased variability associated with small sample sizes. The t-distribution is similar to the normal distribution but has thicker tails, which allows for the uncertainty in estimating the population standard deviation.
Here's a step-by-step breakdown of how a t-test works:
  • Calculate the sample mean and standard deviation.
  • Determine the desired significance level and calculate the critical t-value using a t-distribution table.
  • Compute the t-statistic using the formula: \[ t = \frac{\bar{x} - \mu}{s/\sqrt{n}} \]
  • Compare the calculated t-statistic with the critical t-value to draw conclusions.
By following these steps, we can understand whether our sample data supports or rejects the null hypothesis.
Significance Level
The significance level, often denoted as \( \alpha \), is a crucial part of hypothesis testing, providing the criterion for deciding whether to accept or reject the null hypothesis. In our exercise, we chose a significance level of 0.05, which is a standard choice in many scientific studies. This means that there is a 5% probability of rejecting the null hypothesis when it is actually true (Type I error).
The significance level helps define the "cut-off" threshold or critical value for the t-test. When our test statistic exceeds this critical value, we reject the null hypothesis, implying that there's sufficient evidence to support the alternative hypothesis. The significance level also reflects the confidence we have in our results:
  • Lower significance levels (e.g., 0.01) indicate a higher threshold for rejecting the null hypothesis, thus reducing the risk of a Type I error but making it harder to detect a true effect.
  • Higher significance levels (e.g., 0.10) can increase the risk of a Type I error but make it easier to detect smaller effects.
This delicate balance between sensitivity and specificity is a critical consideration in statistical testing.
Sample Mean
The sample mean, represented as \( \bar{x} \), is a fundamental concept in statistics. It is the average value of a set of observations from a sample and is used to make inferences about the population mean \( \mu \). In hypothesis testing, particularly in t-tests, the sample mean is used to determine whether there is a significant difference from the population mean specified in the null hypothesis.
Calculating the sample mean involves adding together all the sample values and dividing by the number of observations in the sample. For instance, if our sample values were 12, 10, 11, 14, and 9, the sample mean would be:
\[ \bar{x} = \frac{12 + 10 + 11 + 14 + 9}{5} = 11.2 \]
In the problem, the sample mean was given as 12. Since it's higher than the hypothesized population mean of 10, we performed a t-test to determine if this difference was statistically significant. The sample mean provides an estimate of the central tendency of the data and is essential in comparing it to theoretical expectations or known values.
Null Hypothesis
The null hypothesis, denoted as \( H_0 \), is a statement that there is no effect or no difference, and it serves as the starting point for hypothesis testing. It is the hypothesis that researchers typically aim to test or challenge, expecting to find evidence in favor of the alternative hypothesis \( H_1 \).
In our exercise, the null hypothesis was \( H_0: \mu \leq 10 \), implying that the population mean is less than or equal to 10. The goal was to use statistical evidence to determine whether this assumption could be rejected in favor of the alternative hypothesis \( H_1: \mu > 10 \).
Testing the null hypothesis involves:
  • Choosing an appropriate significance level.
  • Determining the necessary test (e.g., t-test) based on the sample size and data type.
  • Calculating the test statistic from the sample data.
  • Comparing the test statistic to a critical value to decide whether to reject \( H_0 \).
Refuting the null hypothesis suggests that there is enough evidence to support the alternative hypothesis. However, failure to reject \( H_0 \) does not prove it true, it simply implies insufficient evidence to support \( H_1 \).

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Most popular questions from this chapter

A United Nations report shows the mean family income for Mexican migrants to the United States is 27,000$ per year. A FLOC (Farm Labor Organizing Committee) evaluation of 25 Mexican family units reveals a mean to be 30,000 with a sample standard deviation of 10,000 . Does this information disagree with the United Nations report? Apply the 0.01 significance level.

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Chicken Delight claims that 90 percent of its orders are delivered within 10 minutes of the time the order is placed. A sample of 100 orders revealed that 82 were delivered within the promised time. At the .10 significance level, can we conclude that less than 90 percent of the orders are delivered in less than 10 minutes?

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