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Chapter 9: Problem 6

A random sample of \(n=35\) observations from a quantitative population produced a mean \(\bar{x}=2.4\) and a standard deviation \(s=.29 .\) Suppose your research objective is to show that the population mean \(\mu\) exceeds 2.3 a. Give the null and alternative hypotheses for the tes b. Locate the rejection region for the test using a \(5 \%\) significance level. c. Find the standard error of the mean. d. Before you conduct the test, use your intuition to decide whether the sample mean \(\bar{x}=2.4\) is likely or unlikely, assuming that \(\mu=2.3 .\) Now conduct the test. Do the data provide sufficient evidence to indicate that \(\mu>2.3 ?\)

Short Answer

Expert verified
Answer: Yes, there is sufficient evidence to support the claim that the population mean is greater than 2.3.

Step by step solution

01

Set up the null and alternative hypotheses

The null hypothesis (H0) states that the population mean is equal to 2.3, while the alternative hypothesis (H1) states it is greater than 2.3. Mathematically, H0: 渭 = 2.3 H1: 渭 > 2.3
02

Determine the rejection region using the 5% significance level

To find the rejection region, we need to find the critical value from the t-distribution (as we don't know the population standard deviation) table using the given sample size (n = 35) and significance level (伪 = 0.05). The degrees of freedom will be n-1 = 34. The critical value we get for 伪=0.05 and df=34 is 1.692. RedirectTo(latex_code_table) The rejection region occurs when the t-value is greater than the critical value: t > 1.692.
03

Calculate the standard error of the mean

The standard error of the mean is calculated as follows: SEM = \(\frac{s}{\sqrt{n}}\) where s is the sample standard deviation and n is the sample size. Plugging in the given values (s = 0.29, n = 35): SEM = \(\frac{0.29}{\sqrt{35}} \approx 0.049\)
04

Intuition and hypothesis testing

Intuitively, the sample mean (2.4) is greater than the hypothesized population mean (2.3). To formally test this, we need to compute the t-value: t = \(\frac{\bar{x} - \mu_{0}}{SEM}\) where \(\bar{x}\) is the sample mean, \(\mu_{0}\) is the hypothesized population mean, and SEM is the standard error of the mean calculated in Step 3. Plugging in the values (2.4, 2.3, 0.049) we get: t = \(\frac{2.4 - 2.3}{0.049} \approx 2.04\) Now, we compare the calculated t-value with the critical value (1.692) obtained from the table in Step 2. The t-value (2.04) is greater than the critical value, so the data falls in the rejection region. Therefore, we reject the null hypothesis (H0) and conclude that there is sufficient evidence to indicate that the population mean (渭) is greater than 2.3.

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Key Concepts

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

Null Hypothesis
The null hypothesis is a foundational concept in hypothesis testing. It is a statement that suggests there is no effect or no difference, often representing a default position to be tested against. In our exercise, the null hypothesis (denoted by \( H_0 \)) proposes that the population mean \( \mu \) is equal to a specified value, in this case, 2.3. This assumption stands unless the evidence strongly suggests otherwise. Understanding the null hypothesis is crucial because it helps us define what we are testing against the collected data. Here鈥檚 a concise breakdown:
  • The null hypothesis is usually expressed as an equation, \( H_0: \mu = 2.3 \).
  • It assumes that any observed difference in sample data is due to random chance.
  • Rejecting the null hypothesis implies that the evidence supports an effect or a difference, while failing to reject it means there鈥檚 insufficient evidence to show a significant change or difference.
Alternative Hypothesis
The alternative hypothesis acts as the opposite of the null hypothesis. It represents the idea that there is an effect, a difference, or that something has changed. In hypothesis testing, we seek evidence to support the alternative hypothesis. For the given problem, the alternative hypothesis (denoted as \( H_1 \) or \( H_a \)) suggests that the population mean \( \mu \) is greater than 2.3. This is particularly relevant in research scenarios where proving a change or effect is the objective. Key points include:
  • The alternative hypothesis can be one-sided or two-sided, depending on what we aim to prove. Here, it鈥檚 one-sided: \( H_1: \mu > 2.3 \).
  • Validating the alternative hypothesis requires statistical evidence strong enough to reject the null hypothesis.
  • If the data shows a significant difference, we lean towards the alternative hypothesis, indicating an effect or change.
Standard Error of the Mean
The standard error of the mean (SEM) is a statistic that reflects how much variation exists in the sample mean estimations if we repeatedly drew samples from the same population. It plays a crucial role in hypothesis testing as it influences the test鈥檚 conclusion about the null hypothesis. For our problem, the SEM is calculated using the formula:
\[ SEM = \frac{s}{\sqrt{n}} \]
where \( s \) is the sample standard deviation, and \( n \) is the sample size. In this case, \( SEM = \frac{0.29}{\sqrt{35}} \approx 0.049 \).
  • The smaller the SEM, the more precise the estimation of the population mean.
  • It is used to compute the t-value, which tells us if the sample mean differs significantly from the population mean.
  • SEM decreases as the sample size increases, reflecting more stable estimates with larger samples.
T-Distribution
The t-distribution is a key aspect of hypothesis testing, especially useful when dealing with small sample sizes or unknown population standard deviations. It shares similarities with the normal distribution but has thicker tails, accommodating the variability seen in small samples. In this exercise, the t-distribution is integral to calculating the critical value and determining the rejection region for the null hypothesis.
  • The t-distribution relies on degrees of freedom, calculated as \( n - 1 \) (34 in this case), to match sample variability.
  • It aids in finding the critical value, which defines the threshold above which we reject the null hypothesis; here, we found this critical value to be 1.692.
  • As sample size increases, the t-distribution approaches a normal distribution, offering familiar ground for interpretation of results.
Significance Level
The significance level, often denoted as \( \alpha \), represents the probability of rejecting the null hypothesis when it is true. It is a predefined threshold that helps us decide whether the findings of our statistical test are "statistically significant." In this exercise, a significance level of 5% (\( \alpha = 0.05 \)) is used.
  • A 5% significance level means there is a 5% chance of committing a Type I error - incorrectly rejecting a true null hypothesis.
  • Lower significance levels reduce Type I error risk but may increase Type II error - failing to reject a false null hypothesis.
  • By comparing the p-value or test statistic against this level, we decide if the results support the alternative hypothesis.
Choosing an appropriate significance level is crucial as it sets the strictness of the test, balancing the probability of making errors and finding true effects.

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Most popular questions from this chapter

9.51 Baby's Sleeping Position Does a baby's sleeping position affect the development of motor skills? In one study, published in the Archives of Pediatric Adolescent Medicine, 343 full-term infants were examined at their 4 -month checkups for various developmental milestones, such as rolling over, grasping a rattle, reaching for an object, and so on. \({ }^{16}\) The baby's predominant sleep position \(-\) either prone (on the stomach) or supine (on the back) or side-was determined by a telephone interview with the parent. The sample results for 320 of the 343 infants for whom information was received are shown here: The researcher reported that infants who slept in the side or supine position were less likely to roll over at the 4 -month checkup than infants who slept primarily in the prone position \((P<.001) .\) Use a large-sample test of hypothesis to confirm or refute the researcher's conclusion.

Treatment versus Control An experiment was conducted to test the effect of a new drug on a viral infection. After the infection was induced in 100 mice, the mice were randomly split into two groups of \(50 .\) The first group, the control group, received no treatment for the infection, and the second group received the drug. After a 30 -day period, the proportions of survivors, \(\hat{p}_{1}\) and \(\hat{p}_{2}\), in the two groups were found to be .36 and \(.60,\) respectively. a. Is there sufficient evidence to indicate that the drug is effective in treating the viral infection? Use \(\alpha=05\) b. Use a \(95 \%\) confidence interval to estimate the actual difference in the survival rates for the treated versus the control groups.

Suppose you wish to detect a difference between \(\mu_{1}\) and \(\mu_{2}\) (either \(\mu_{1}>\mu_{2}\) or \(\mu_{1}<\mu_{2}\) ) and, instead of running a two-tailed test using \(\alpha=.05,\) you use the following test procedure. You wait until you have collected the sample data and have calculated \(\bar{x}_{1}\) and \(\bar{x}_{2}\). If \(\bar{x}_{1}\) is larger than \(\bar{x}_{2},\) you choose the alternative hypothesis \(H_{\mathrm{a}}: \mu_{1}>\mu_{2}\) and run a one-tailed test placing \(\alpha_{1}=.05\) in the upper tail of the \(z\) distribution. If, on the other hand, \(\bar{x}_{2}\) is larger than \(\bar{x}_{1},\) you reverse the procedure and run a one-tailed test, placing \(\alpha_{2}=.05\) in the lower tail of the \(z\) distribution. If you use this procedure and if \(\mu_{1}\) actually equals \(\mu_{2},\) what is the probability \(\alpha\) that you will conclude that \(\mu_{1}\) is not equal to \(\mu_{2}\) (i.e., what is the probability \(\alpha\) that you will incorrectly reject \(H_{0}\) when \(H_{0}\) is true)? This exercise demonstrates why statistical tests should be formulated prior to observing the data.

Potency of an Antibiotic A drug manufacturer claimed that the mean potency of one of its antibiotics was \(80 \%\). A random sample of \(n=100\) capsules were tested and produced a sample mean of \(\bar{x}=79.7 \%\) with a standard deviation of \(s=.8 \% .\) Do the data present sufficient evidence to refute the manufacturer's claim? Let \(\alpha=.05 .\) a. State the null hypothesis to be tested. b. State the alternative hypothesis. c. Conduct a statistical test of the null hypothesis and state your conclusion.

A random sample of \(n=1400\) observations from a binomial population produced \(x=529 .\) a. If your research hypothesis is that \(p\) differs from .4, what hypotheses should you test? b. Calculate the test statistic and its \(p\) -value. Use the \(p\) -value to evaluate the statistical significance of the results at the \(1 \%\) level. c. Do the data provide sufficient evidence to indicate that \(p\) is different from \(.4 ?\)
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