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Chapter 10: Problem 10

To test \(H_{0}: \mu=45\) versus \(H_{1}: \mu \neq 45,\) a simple random sample of size \(n=40\) is obtained. (a) Does the population have to be normally distributed to test this hypothesis by using the methods presented in this section? Why? (b) If \(\bar{x}=48.3\) and \(s=8.5,\) compute the test statistic. (c) Draw a \(t\) -distribution with the area that represents the P-value shaded. (d) Approximate and interpret the \(P\) -value. (e) If the researcher decides to test this hypothesis at the \(\alpha=0.01\) level of significance, will the researcher reject the null hypothesis? Why? (f) Construct a \(99 \%\) confidence interval to test the hypothesis.

Short Answer

Expert verified
a) No, due to CLT; b) t 鈮� 2.456; c) Shaded areas beyond 卤2.456; d) P 鈮� 0.019; e) Do not reject H0; f) CI: (44.654, 51.946).

Step by step solution

01

Title - Determine if population must be normally distributed

If the sample size n is large (typically n 鈮� 30), the Central Limit Theorem states that the sampling distribution of the sample mean is approximately normal, regardless of the population distribution. Here, n = 40, which is greater than 30.
02

Title - Compute the test statistic

To compute the test statistic, use the formula for the t-test:\[ t = \frac{\bar{x} - \mu}{s/\sqrt{n}} \]Plugging in the values:\[ t = \frac{48.3 - 45}{8.5/\sqrt{40}} \]\[ t = \frac{3.3}{1.344} \approx 2.456 \]
03

Title - Draw t-distribution with P-value area shaded

Draw a t-distribution curve with degrees of freedom, df = n - 1 = 40 - 1 = 39. Shade the two tail regions that correspond to the calculated test statistic value of 卤2.456.
04

Title - Approximate and interpret the P-value

The P-value is the probability that the test statistic would be at least as extreme as the observed value, assuming the null hypothesis is true. Using a t-table or calculator, find the P-value for t = 卤2.456 with 39 degrees of freedom. The P-value is approximately 0.019. This means there's a 1.9% chance of observing a test statistic as extreme as 卤2.456 if the null hypothesis is true.
05

Title - Decision at 伪 = 0.01 level of significance

Compare the P-value (0.019) with the significance level (0.01). Since 0.019 > 0.01, the P-value is greater than the significance level. Thus, do not reject the null hypothesis.
06

Title - Construct 99% confidence interval

Calculate the 99% confidence interval for the mean. Use the formula:\[ \bar{x} \pm t_{\alpha/2} \times \frac{s}{\sqrt{n}} \]Find the critical value t_{\alpha/2} for 39 degrees of freedom. For a 99% confidence level, t_{0.005} 鈮� 2.708. Calculate the margin of error:\[ ME = 2.708 \times \frac{8.5}{\sqrt{40}} \approx 3.646 \]Construct the interval:\[ 48.3 \pm 3.646 \]\[ (44.654, 51.946) \]Since 45 is within this interval, do not reject the null hypothesis.

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Key Concepts

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

Central Limit Theorem
The Central Limit Theorem (CLT) is a fundamental concept in statistics. It states that when you have a sufficiently large sample size (usually n 鈮� 30) from any population with a finite level of variance, the sampling distribution of the sample mean will be approximately normally distributed.

This holds true regardless of the original population distribution.

In the exercise, the sample size n = 40 is greater than 30, meaning that CLT applies. This allows us to use the normal distribution to draw valid conclusions even if the original population isn't normally distributed.
t-test
A t-test is used to determine if there is a significant difference between the means of two groups, which may be related in certain features.

In hypothesis testing, the t-test helps in finding out if we should reject the null hypothesis.

The formula for the t-test in the context of a mean is:
\[ t = \frac{\bar{x} - \text{\textmu}}{s/\text{\textsqrt}{n}} \]
Here:
  • \( \bar{x} \) is the sample mean
  • \( \text{\textmu} \) is the population mean
  • \( s \) is the sample standard deviation
  • \( n \) is the sample size

For the given exercise, the values are \( \bar{x} = 48.3 \), \( \text{\textmu} = 45 \), \( s = 8.5 \), and \( n = 40 \). Plugging these into the formula gives a t value of approximately 2.456.
p-value
The p-value in hypothesis testing indicates the probability of obtaining test results at least as extreme as the results actually observed, under the assumption that the null hypothesis is correct.

A smaller p-value means that the observed data is unlikely under the null hypothesis.

In the exercise, the p-value for t = 卤2.456 with 39 degrees of freedom is approximately 0.019. This means there is a 1.9% chance of getting a test statistic as extreme as 卤2.456 if the null hypothesis is true.

Understanding the p-value helps indicate how probable your test results are purely by chance.
Confidence Interval
A confidence interval provides a range of values that is likely to contain the population parameter of interest.

It is expressed as: \[ \bar{x} \text{\textpm} (t_{\text{\textalpha}/2} \times \frac{s}{\text{\textsqrt}{n}})\]
Here:
  • \( \bar{x} \) is the sample mean
  • \( t_{\text{\textalpha}/2} \) is the t value for the desired confidence level
  • \( s \) is the sample standard deviation
  • \( n \) is the sample size

In the exercise, to construct a 99% confidence interval, we use a critical value of approximately 2.708 for 39 degrees of freedom. The resulting confidence interval is (44.654, 51.946), meaning we are 99% confident that the true mean lies within this range.
Level of Significance
The level of significance (\( \text{\textalpha} \)) is the threshold we use to decide whether to reject the null hypothesis.

Common significance levels are 0.05, 0.01, and 0.10. In this case, the significance level is 0.01, meaning there is a 1% risk of concluding that a difference exists when there is no actual difference.

This decision rule states if the p-value is less than \( \text{\textalpha} \), you reject the null hypothesis. Otherwise, you do not reject it.

For the exercise's p-value of 0.019, it is greater than the 0.01 significance level. Therefore, the null hypothesis is not rejected.

The level of significance thus sets the criteria for making this critical decision.

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Most popular questions from this chapter

Professors Honey Kirk and Diane Lerma of Palo Alto College developed a "learning community curriculum that blended the developmental mathematics and the reading curriculum with a structured emphasis on study skills." In a typical developmental mathematics course at Palo Alto College, \(50 \%\) of the students complete the course with a letter grade of \(\mathrm{A}, \mathrm{B},\) or \(\mathrm{C} .\) In the experimental course, of the 16 students enrolled, 11 completed the course with a letter grade of \(\mathrm{A}, \mathrm{B},\) or \(\mathrm{C} .\) Do you believe the experimental course was effective at the \(\alpha=0.05\) level of significance? (a) State the appropriate null and alternative hypotheses. (b) Verify that the normal model may not be used to estimate the P-value (c) Explain why this is a binomial experiment. (d) Determine the \(P\) -value using the binomial probability distribution. State your conclusion to the hypothesis test. (e) Suppose the course is taught with 48 students and 33 complete the course with a letter grade of \(\mathrm{A}, \mathrm{B},\) or \(\mathrm{C}\). Verify the normal model may now be used to estimate the \(P\) -value. (f) Use the normal model to obtain and interpret the \(P\) -value. State your conclusion to the hypothesis test. (g) Explain the role that sample size plays in the ability to reject statements in the null hypothesis.

Throwing darts at the stock pages to decide which companies to invest in could be a successful stock-picking strategy. Suppose a researcher decides to test this theory and randomly chooses 100 companies to invest in. After 1 year, 53 of the companies were considered winners; that is, they outperformed other companies in the same investment class. To assess whether the dart-picking strategy resulted in a majority of winners, the researcher tested \(H_{0}: p=0.5\) versus \(H_{1}: p>0.5\) and obtained a \(P\) -value of \(0.2743 .\) Explain what this \(P\) -value means and write a conclusion for the researcher.

Nexium is a drug that can be used to reduce the acid produced by the body and heal damage to the esophagus due to acid reflux. The manufacturer of Nexium claims that more than \(94 \%\) of patients taking Nexium are healed within 8 weeks. In clinical trials, 213 of 224 patients suffering from acid reflux disease were healed after 8 weeks. Test the manufacturer's claim at the \(\alpha=0.01\) level of significance.

Suppose we are testing the hypothesis \(H_{0}: p=0.65\) versus \(H_{1}: p \neq 0.65\) and we find the \(P\) -value to be \(0.02 .\) Explain what this means. Would you reject the null hypothesis? Why?

A can of soda is labeled as containing 12 fluid ounces. The quality control manager wants to verify that the filling machine is neither over-filling nor under-filling the cans. (a) Determine the null and alternative hypotheses that would be used to determine if the filling machine is calibrated correctly. (b) The quality control manager obtains a sample of 75 cans and measures the contents. The sample evidence leads the manager to reject the null hypothesis. Write a conclusion for this hypothesis test. (c) Suppose, in fact, the machine is not out of calibration. Has a Type I or Type II error been made? (d) Management has informed the quality control department that it does not want to shut down the filling machine unless the evidence is overwhelming that the machine is out of calibration. What level of significance would you recommend the quality control manager use? Explain.
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