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Chapter 9: Problem 39

A Cure for Insomnia An experimenter has prepared a drug-dose level that he claims will induce sleep for at least \(80 \%\) of people suffering from insomnia. After examining the dosage we feel that his claims regarding the effectiveness of his dosage are inflated. In an attempt to disprove his claim, we administer his prescribed dosage to 50 insomniacs and observe that 37 of them have had sleep induced by the drug dose. Is there enough evidence to refute his claim at the \(5 \%\) level of significance?

Short Answer

Expert verified
Answer: Yes, we have enough evidence to refute the experimenter's claim, as the p-value (0.033) is less than the 5% significance level, indicating that the proportion of insomniacs who sleep with the given drug dosage is significantly less than 80%.

Step by step solution

01

State the null and alternative hypothesis

The null hypothesis (H0) asserts that the claim made by the experimenter is true, i.e., the proportion of sleep induced is equal to or greater than 80% (\(\ge\) 0.8). H0: p \(\ge\) 0.8 The alternative hypothesis (Ha) is that there is evidence against the experimenter's claim. The proportion of sleep induced is less than 80% (< 0.8). Ha: p < 0.8
02

Calculate the test statistic

We will use the given sample data to calculate the test statistic z for a one-sample proportion hypothesis test: \(z = \frac{(\hat{p} - p_0)}{\sqrt{\frac{p_0(1 - p_0)}{n}}}\) where \(\hat{p}\) is the sample proportion, \(p_0\) is the claimed proportion, and n is the sample size. Using the given data, \(\hat{p} = \frac{37}{50} = 0.74\), \(p_0 = 0.8\), and \(n = 50\). Substituting these values, we can calculate the value of z: \(z = \frac{(0.74 - 0.8)}{\sqrt{\frac{0.8(1 - 0.8)}{50}}}\) \(z \approx -1.837\)
03

Calculate the p-value

The p-value represents the probability of obtaining the test statistic or a more extreme result under the null hypothesis. Since the alternative hypothesis is one-sided (p < 0.8), we will calculate the p-value using the standard normal table or a calculator as follows: p-value = P(Z < -1.837) p-value \(\approx\) 0.033
04

Compare the p-value with the significance level

Now, we will compare the p-value with the given significance level (伪 = 0.05) to determine if there is enough evidence to reject the null hypothesis: If p-value 鈮� 伪, then we reject the null hypothesis in favor of the alternative hypothesis. If p-value > 伪, then we fail to reject the null hypothesis. In this case, the p-value (0.033) is less than the significance level (0.05), so we reject the null hypothesis.
05

Conclusion

As we reject the null hypothesis, there is enough evidence to conclude that the proportion of insomniacs who sleep with the given drug dosage is significantly less than 80%. Therefore, we have strong evidence to refute the experimenter's claim regarding the effectiveness of his dosage at the 5% level of significance.

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Key Concepts

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

Null Hypothesis
In hypothesis testing, the null hypothesis is the starting point for any statistical test. It represents a default position or a statement that there is no effect or no difference. In this context, the null hypothesis (\(H_0\)) asserts that the experimenter's claim is correct, meaning the proportion of people who experience sleep after taking the drug is at least 80%.

The null hypothesis plays a vital role because it sets the stage for examination, asking whether the observed data provide strong enough evidence to reject this assumption. When we say "reject the null hypothesis," we're suggesting the data showed us something unexpected when assuming the null hypothesis was true.
  • Symbolically, \(H_0: p \geq 0.8\) where \(p\) denotes the true proportion of patients affected by the drug.
  • It's important to note that failing to reject the null hypothesis does not prove it to be true; it simply means there wasn't enough evidence against it.
Alternative Hypothesis
The alternative hypothesis (\(H_a\)) is what researchers typically aim to support with evidence. It represents a statement contradicting the null hypothesis. In this experiment about curing insomnia with a drug, the alternative hypothesis posits that the proportion of insomniacs for whom the drug induces sleep is less than 80%.

The alternative hypothesis is crucial because it formulates what we're testing for or attempting to demonstrate as potentially true, based on the data collected.
  • For the insomniacs experiment, \(H_a: p < 0.8\) is set up to challenge the experimenter's inflated claim.
  • If the statistical test results lead us to reject the null, this indirectly suggests our alternative hypothesis has merit.
P-value
The p-value helps us decide whether the observed data are consistent with the null hypothesis. It measures the probability of obtaining the test statistic or something more extreme, assuming that the null hypothesis is true.

In essence, the smaller this value, the more surprising the data are if the null hypothesis really was true.
  • In our insomnia study, the p-value dropped to around 0.033, indicating that if the drug truly worked for at least 80% of people, the observed result of only 74% effectiveness is somewhat unlikely.
  • A low p-value (<0.05 in many studies) can suggest that there's significant evidence against the null hypothesis.
  • Keep in mind that the p-value alone isn't proof. It should contribute to the overall decision in conjunction with other study design elements.
Significance Level
The significance level (\(\alpha\)), often denoted by symbols like 0.05, sets the cutoff for making decisions based on the p-value. It's the threshold for what counts as "statistically significant" and indicates the risk you are willing to accept for rejecting a true null hypothesis.

Having a preset significance level offers a consistent benchmark.
  • In our scenario, the significance level is \(0.05\) (5%). It tells us that we are okay with a 5% chance of incorrectly rejecting the null hypothesis.
  • When the p-value is less than the significance level, like it was here (0.033 < 0.05), it can be interpreted that there's enough evidence to discard the null hypothesis in favor of the alternative.
  • Setting an appropriate significance level before conducting your test will help maintain the transparency and integrity of your hypothesis testing.

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Most popular questions from this chapter

Taste Testing In a head-to-head taste test of store-brand foods versus national brands, Consumer Reports found that it was hard to find a taste difference in the two. \({ }^{10}\) If the national brand is indeed better than the store brand, it should be judged as better more than \(50 \%\) of the time. a. State the null and alternative hypothesis to be tested. Is this a one- or a two-tailed test? b. Suppose that, of the 35 food categories used for the taste test, the national brand was found to be better than the store brand in eight of the taste comparisons. Use this information to test the hypothesis in part a. Use \(\alpha=.01\). What practical conclusions can you draw from the results?

Independent random samples of 280 and 350 observations were selected from binomial populations 1 and 2 , respectively. Sample 1 had 132 successes, and sample 2 had 178 successes. Do the data present sufficient evidence to indicate that the proportion of successes in population 1 is smaller than the proportion in population \(2 ?\) Use one of the two methods of testing presented in this section, and explain your conclusions.

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.

Landlines Passe? According to a new nationwide survey from the Pew Research Center's Social \& Demographic Trends project, \(62 \%\) of Americans consider a landline phone a necessity of life, down from \(68 \%\) last year, and only \(43 \%\) of Americans consider a TV a necessity. \({ }^{11}\) Both questions were on the second of two forms used in the survey and involved \(n\) \(=1483\) individuals contacted by either a landline phone or a cell phone. Use a test of hypothesis to determine whether the \(\hat{p}=.62\) figure is a significant drop from last year's value of \(p=.68\) with \(\alpha=.01\). What conclusions can you draw from this analysis?

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 \(\mu>2.3 ?\)
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