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Chapter 4: Problem 160

Hypotheses for a statistical test are given, followed by several possible confidence intervals for different samples. In each case, use the confidence interval to state a conclusion of the test for that sample and give the significance level used. Hypotheses: \(H_{0}: \mu_{1}=\mu_{2}\) vs \(H_{a}: \mu_{1} \neq \mu_{2} .\) In addition, in each case for which the results are significant, state which group ( 1 or 2 ) has the larger mean. (a) \(95 \%\) confidence interval for \(\mu_{1}-\mu_{2}\) : 0.12 to 0.54 (b) \(99 \%\) confidence interval for \(\mu_{1}-\mu_{2}\) : -2.1 to 5.4 (c) \(90 \%\) confidence interval for \(\mu_{1}-\mu_{2}\) : -10.8 to -3.7

Short Answer

Expert verified
Based on the given confidence intervals, we reject the null hypothesis for samples (a) and (c) and fail to reject it for sample (b). In case (a), group 1 has a larger mean. In case (b), we cannot determine which group has a larger mean. In case (c), group 2 has a larger mean.

Step by step solution

01

Interpret Confidence Interval (a)

A 95% confidence interval for \( \mu_{1} - \mu_{2} \) ranges from 0.12 to 0.54. This interval excludes 0, hence we reject the null hypothesis \( H_0: \mu_{1} = \mu_{2} \) at the 5% significance level. The positive difference suggests that group 1 has a larger mean.
02

Interpret Confidence Interval (b)

A 99% confidence interval for \( \mu_{1} - \mu_{2} \) ranges from -2.1 to 5.4. This interval contains 0, hence we fail to reject the null hypothesis \( H_0: \mu_{1} = \mu_{2} \) at the 1% significance level. We cannot conclude which group has a larger mean.
03

Interpret Confidence Interval (c)

A 90% confidence interval for \( \mu_{1} - \mu_{2} \) ranges from -10.8 to -3.7. This interval also excludes 0; hence we reject the null hypothesis \( H_0: \mu_{1} = \mu_{2} \) at the 10% significance level. The negative difference suggests that group 2 has a larger mean.

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Key Concepts

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

Confidence Interval Interpretation
When you're interpreting a confidence interval, you're essentially trying to understand the range within which you can expect your statistical parameter to fall.
  • A confidence interval provides an estimated range derived from a sample dataset.
  • It shows where you might find the true difference between the means in the whole population if you sampled the population over and over again.
For instance, if a 95% confidence interval for \( \mu_1 - \mu_2 \) ranges from 0.12 to 0.54, it means that we are 95% confident that the true mean difference is somewhere between these numbers. Crucially, if a confidence interval doesn't include 0,it suggests that there is a significant difference between the means of the two groups. In other words, the null hypothesis can often be rejected, indicating that a real effect or difference exists between the groups. Conversely, if 0 is inside the interval, it implies there's no significant difference, as seen in example (b) where the interval was -2.1 to 5.4.Here, we conclude that we cannot reject the null hypothesis.
Significance Level
The significance level, denoted by \( \alpha \), is a threshold set by the researcher which determines how strong the evidence must be to reject the null hypothesis. It's foundational in hypothesis testing because:
  • It reflects our tolerance for making a Type I error: erroneously rejecting a true null hypothesis.
  • A lower significance level implies stricter criteria for rejecting the null hypothesis.
  • Common levels of significance are 0.05, 0.01, and 0.10, corresponding to confidence levels of 95%, 99%, and 90% respectively.
In this context, if our 95% confidence interval for \( \mu_1 - \mu_2 \) is between 0.12 and 0.54,we reject the null hypothesis at a significance level of 5% (\( \alpha = 0.05 \)).It implies that there is less than a 5% probability that the results observed are due to random chance.Significance levels critically aid us in determining whether the observed effects are "real" or merely the consequence of random sampling error.
Null and Alternative Hypotheses
To conduct a statistical test, one starts by defining the null and alternative hypotheses. These form the cornerstone of any hypothesis testing framework.
  • The **null hypothesis** (\( H_0 \): \( \mu_1 = \mu_2 \)) posits that there is no difference between the population means.
  • The **alternative hypothesis** (\( H_a \): \( \mu_1 eq \mu_2 \)) suggests there is some difference.
The null hypothesis serves as the default or status quo, while the alternative hypothesis is what the researcher aims to support.In practice, we use data to test whether we can confidently refute the null hypothesis in favor of the alternative. When the confidence interval for the difference \( \mu_1 - \mu_2 \) excludes zero, we have reason to reject the null hypothesis and thus infer a difference.When it includes zero, however, as in interval case (b) from -2.1 to 5.4, we lack the statistical evidence to reject the null hypothesis, indicating that any observed difference could be due to chance rather than an actual effect.

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Most popular questions from this chapter

A study suggests that exposure to UV rays through the car window may increase the risk of skin cancer. \(^{52}\) The study reviewed the records of all 1,050 skin cancer patients referred to the St. Louis University Cancer Center in 2004\. Of the 42 patients with melanoma, the cancer occurred on the left side of the body in 31 patients and on the right side in the other 11 . (a) Is this an experiment or an observational study? (b) Of the patients with melanoma, what proportion had the cancer on the left side? (c) A bootstrap \(95 \%\) confidence interval for the proportion of melanomas occurring on the left is 0.579 to \(0.861 .\) Clearly interpret the confidence interval in the context of the problem. (d) Suppose the question of interest is whether melanomas are more likely to occur on the left side than on the right. State the null and alternative hypotheses. (e) Is this a one-tailed or two-tailed test? (f) Use the confidence interval given in part (c) to predict the results of the hypothesis test in part (d). Explain your reasoning. (g) A randomization distribution gives the p-value as 0.003 for testing the hypotheses given in part (d). What is the conclusion of the test in the context of this study? (h) The authors hypothesize that skin cancers are more prevalent on the left because of the sunlight coming in through car windows. (Windows protect against UVB rays but not UVA rays.) Do the data in this study support a conclusion that more melanomas occur on the left side because of increased exposure to sunlight on that side for drivers?

For each situation described, indicate whether it makes more sense to use a relatively large significance level (such as \(\alpha=0.10\) ) or a relatively small significance level (such as \(\alpha=0.01\) ). Testing a new drug with potentially dangerous side effects to see if it is significantly better than the drug currently in use. If it is found to be more effective, it will be prescribed to millions of people.

The null and alternative hypotheses for a test are given as well as some information about the actual sample(s) and the statistic that is computed for each randomization sample. Indicate where the randomization distribution will be centered. In addition, indicate whether the test is a left-tail test, a right-tail test, or a twotailed test. Hypotheses: \(H_{0}: \mu=10\) vs \(H_{a}: \mu>10\) Sample: \(\bar{x}=12, s=3.8, n=40\)

Figure 4.25 shows a scatterplot of the acidity (pH) for a sample of \(n=53\) Florida lakes vs the average mercury level (ppm) found in fish taken from each lake. The full dataset is introduced in Data 2.4 on page 71 and is available in FloridaLakes. There appears to be a negative trend in the scatterplot, and we wish to test whether there is significant evidence of a negative association between \(\mathrm{pH}\) and mercury levels. (a) What are the null and alternative hypotheses? (b) For these data, a statistical software package produces the following output: $$ r=-0.575 \quad p \text { -value }=0.000017 $$ Use the p-value to give the conclusion of the test. Include an assessment of the strength of the evidence and state your result in terms of rejecting or failing to reject \(H_{0}\) and in terms of \(\mathrm{pH}\) and mercury. (c) Is this convincing evidence that low \(\mathrm{pH}\) causes the average mercury level in fish to increase? Why or why not?

In Exercises 4.40 to 4.44 , null and alternative hypotheses for a test are given. Give the notation \((\bar{x},\) for example) for a sample statistic we might record for each simulated sample to create the randomization distribution. \(H_{0}: p=0.5\) vs \(H_{a}: p \neq 0.5\)
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