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Given the following hypothesis: $$\begin{array}{l}H_{0}: \mu \geq 20 \\\H_{1}: \mu<20\end{array}$$ A random sample of five resulted in the following values: \(18,15,12,19,\) and \(21 .\) Using the .01 significance level, can we conclude the population mean is less than \(20 ?\) a. State the decision rule. b. Compute the value of the test statistic. c. What is your decision regarding the null hypothesis? d. Estimate the \(p\) -value.

Short Answer

Expert verified
Fail to reject the null hypothesis; the population mean is not significantly less than 20 at the 0.01 level.

Step by step solution

01

Understand the Decision Rule

The decision rule for a one-sample t-test is based on the critical value of t from the t-distribution table. With \( n = 5 \), the degrees of freedom (df) is \( n-1 = 4 \). For a one-tailed test at a 0.01 significance level, find the critical t-value \( t_c \). If the calculated t-value is less than \( t_c \), reject the null hypothesis \( H_0 \).
02

Calculate Sample Mean and Standard Deviation

Find the sample mean \( \bar{x} \) and standard deviation \( s \) from the sample values. The values are 18, 15, 12, 19, and 21. Calculate:\[ \bar{x} = \frac{18 + 15 + 12 + 19 + 21}{5} = 17 \]To calculate the standard deviation: \[ s = \sqrt{\frac{(18-17)^2 + (15-17)^2 + (12-17)^2 + (19-17)^2 + (21-17)^2}{5-1}} \approx 3.46 \]
03

Compute the Test Statistic

Use the formula for the one-sample t-test statistic:\[ t = \frac{\bar{x} - \mu_0}{s/\sqrt{n}} \]Substitute the known values:\[ t = \frac{17 - 20}{3.46/\sqrt{5}} \approx -1.94 \]
04

Find Critical Value and Make Decision

Using the t-distribution table at 4 degrees of freedom and 0.01 significance level for a one-tailed test, find the critical value \( t_c \) which is approximately -3.747. Compare this with the calculated t-value (-1.94). Since -1.94 is greater than -3.747, we fail to reject the null hypothesis \( H_0 \).
05

Estimate the p-value

The p-value for the calculated t-value (-1.94) at 4 degrees of freedom can be found using a t-table or statistical software. It is greater than 0.01, indicating that there is not enough evidence to reject the null hypothesis \( H_0 \) at the 0.01 level.

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Key Concepts

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

One-sample t-test
The one-sample t-test allows you to determine if a sample mean is significantly different from a known value, often a population mean.
It is useful when comparing the means of a single group against a theoretical value.
Here's how it works:
  • Take a sample from your data set. For instance, consider the sample values: 18, 15, 12, 19, and 21.
  • Calculate the sample mean and standard deviation.
  • Then, compute the t-test statistic, which assesses how far the sample mean is from the population mean, scaled by the variability of the sample data.
In the exercise, we used the formula:\[t = \frac{\bar{x} - \mu_0}{s/\sqrt{n}}\]Where:- \( \bar{x} \) is the sample mean- \( \mu_0 \) is the hypothesized population mean (20 in this case)- \( s \) is the sample standard deviation- \( n \) is the sample sizeThis test assumes that the sample data is approximately normally distributed and is particularly useful when dealing with smaller samples.
Significance Level
The significance level, often denoted by \( \alpha \), is the threshold used to determine when to reject the null hypothesis.
A common choice is 0.05, but in our case, we use a stricter level of 0.01, indicating a high confidence requirement.
What does this mean?
  • It's the probability of rejecting the null hypothesis when it is actually true (Type I error).
  • Lower significance levels reflect higher standards for evidence against the null hypothesis.
In practice:- A significance level of 0.01 means there's only a 1% risk of mistakenly rejecting the null hypothesis.This standard helps ensure the results are not due to random chance, and the chosen significance level should reflect the consequences of error in decision-making.
Null Hypothesis
The null hypothesis (\( H_0 \)) is a statement that there is no effect or no difference, and it is the default position in hypothesis testing.
In the context of the exercise, the null hypothesis is that the population mean \( \mu \) is greater than or equal to 20.
Why is it important?
  • The null hypothesis serves as a baseline that helps us determine if the observed data is significantly different.
  • It is what we assume to be true unless we have evidence to prove otherwise.
  • Testing the null hypothesis involves conducting a t-test and comparing the resulting p-value to the chosen significance level.
Only when evidence against \( H_0 \) is strong enough, do we reject it in favor of an alternative hypothesis (\( H_1 \)). It helps provide a structured approach to statistical inference.
p-value
The p-value is a measure that helps you determine the significance of your results in hypothesis testing.
It tells us how likely it is to observe data at least as extreme as ours under the null hypothesis \( H_0 \).
Understanding the p-value:
  • It ranges from 0 to 1. A smaller p-value means stronger evidence against \( H_0 \).
  • It's compared to the significance level \( \alpha \) to decide whether to reject \( H_0 \).
In detail:- A p-value less than \( \alpha \) means you reject \( H_0 \), supporting the alternative hypothesis.- In our exercise, the estimated p-value was more than 0.01, which was not sufficient to reject \( H_0 \).The p-value gives a tangible measure of how extreme the sample results are, considering the null hypothesis is true. It's a crucial part of statistical hypothesis testing, providing insight into the determination of the research outcome.

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Most popular questions from this chapter

A recent article in The Wall Street Journal reported that the 30-year mortgage rate is now less than 6 percent. A sample of eight small banks in the Midwest revealed the following 30-year rates (in percent): $$\begin{array}{|llllllll|}\hline 4.8 & 5.3 & 6.5 & 4.8 & 6.1 & 5.8 & 6.2 & 5.6 \\\\\hline\end{array}$$ At the .01 significance level, can we conclude that the 30 -year mortgage rate for small banks is less than 6 percent? Estimate the \(p\) -value.

According to the Coffee Research Organization (http://www.coffeeresearch.org) the typical American coffee drinker consumes an average of 3.1 cups per day. A sample of 12 senior citizens revealed they consumed the following amounts of coffee, reported in cups, yesterday. $$\begin{array}{|llllllllllll|}\hline 3.1 & 3.3 & 3.5 & 2.6 & 2.6 & 4.3 & 4.4 & 3.8 & 3.1 & 4.1 & 3.1 & 3.2 \\\\\hline\end{array}$$ At the .05 significance level does this sample data suggest there is a difference between the national average and the sample mean from senior citizens?

(a) Is this a one- or two-tailed test? (b) What is the decision rule? (c) What is the value of the test statistic? (d) What is your decision regarding \(H_{0} ?\) (e) What is the \(p\) -value? Interpret it. A sample of 36 observations is selected from a normal population. The sample mean is \(21,\) and the population standard deviation is \(5 .\) Conduct the following test of hypothesis using the .05 significance level. $$\begin{array}{l}H_{0}: \mu \leq 20 \\\H_{1}: \mu>20\end{array}$$

Listed below is the rate of return for one year (reported in percent) for a sample of 12 mutual funds that are classified as taxable money market funds $$\begin{array}{|lllllllllll|}\hline 4.63 & 4.15 & 4.76 & 4.70 & 4.65 & 4.52 & 4.70 & 5.06 & 4.42 & 4.51 & 4.24 & 4.52 \\\\\hline\end{array}$$ Using the .05 significance level is it reasonable to conclude that the mean rate of return is more than 4.50 percent?

A spark plug manufacturer claimed that its plugs have a mean life in excess of 22,100 miles. Assume the life of the spark plugs follows the normal distribution. A fleet owner purchased a large number of sets. A sample of 18 sets revealed that the mean life was 23,400 miles and the standard deviation was 1,500 miles. Is there enough evidence to substantiate the manufacturer's claim at the .05 significance level?

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