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Exercises 4.117 to 4.122 give null and alternative hypotheses for a population proportion, as well as sample results. Use StatKey or other technology to generate a randomization distribution and calculate a p-value. StatKey tip: Use "Test for a Single Proportion" and then "Edit Data" to enter the sample information. Hypotheses: \(H_{0}: p=0.6\) vs \(H_{a}: p>0.6\) Sample data: \(\hat{p}=52 / 80=0.65\) with \(n=80\)

Short Answer

Expert verified
After setting up your hypotheses and calculating the sample proportion, go onto technology such as StatKey to generate a randomization distribution based on the data and calculate the p-value. Following this, make a conclusion on the null and alternative hypotheses based on the p-value.

Step by step solution

01

Set up the Null and Alternative Hypotheses

The Null Hypothesis \(H_{0}: p=0.6\) represents the initial claim about a population proportion, stating that the population proportion is 0.6. The Alternative Hypothesis \(H_{a}: p > 0.6\) represents a new claim made about the population proportion stating that the population proportion is greater than 0.6.
02

Calculate the Sample Proportion

The given sample data is 52 out of 80 observations producing a sample proportion \(\hat{p}=52 / 80=0.65\).
03

Generate a Randomization Distribution and Calculate the p-value

After setting up your hypotheses and calculating the sample proportion, use technology such as StatKey to generate a randomization distribution based on \(\hat{p}=0.65\) and calculate the p-value. The p-value is the probability of obtaining a result as extreme or more extreme than the observed data under the null hypothesis.
04

Make a Conclusion about the Hypotheses

If the calculated p-value is less than the significance level (typically 0.05), the null hypothesis will be rejected in favor of the alternative hypothesis. If the p-value is greater than the significance level, there will not be enough evidence to reject the null hypothesis.

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Key Concepts

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

P-Value Calculation
In the realm of hypothesis testing in statistics, the concept of a p-value is pivotal. It quantifies the probability of observing a sample statistic as extreme as the one obtained given that the null hypothesis is true. It's a tool used to determine whether the observed data significantly deviates from what was expected under the null hypothesis.

When using technology like StatKey to calculate the p-value, it's important to enter the sample data correctly. In the exercise provided, with a sample proportion (denoted as \(\hat{p}\)) of 0.65, the software compares this value against a distribution of sample proportions expected if the null hypothesis were true (in this case, that the true population proportion \(p\) is 0.6).

The output from such software typically illustrates how unusual the observed sample is compared to this distribution. If the sample falls far into the tail, the p-value will be small, suggesting that the observed sample may be due to something other than chance, potentially leading us to consider the alternative hypothesis that \(p > 0.6\).
Population Proportion
Population proportion, represented by the symbol \(p\), refers to the share of a certain characteristic or attribute within the entire population. In hypothesis testing, we're often interested in making inferences about this population proportion based on sample data.

To ensure accurate results, it's crucial that the sample is representative of the population. The accuracy of hypothesis tests relies heavily on the assumption that the sample proportion (\(\hat{p}\)) is a reliable estimate of the true population proportion. Our exercise uses a sample size of 80, with 52 showing the characteristic of interest, leading to a sample proportion of 0.65. When we compare this sample proportion to the hypothesized population proportion of 0.6, we assess whether this discrepancy might be due to random variation in sample selection or if it indicates something more substantive about the population.
Randomization Distribution
A randomization distribution is a fundamental concept in understanding how we assess the probability of obtaining our sample results under the null hypothesis. It is a theoretical distribution of outcomes that we might observe if we were able to repeatedly sample from the population, while assuming the null hypothesis is true.

The randomization distribution in our exercise is centered around the null hypothesis value of 0.6. Using software to simulate this process, we produce many hypothetical samples, calculate the proportion for each, and compile these to see the distribution of sample proportions that could arise just by chance. This distribution helps us determine how unusual our actual sample proportion of 0.65 is in the context of the null hypothesis.
Null and Alternative Hypotheses
The core of hypothesis testing is framed by two competing statements: the null hypothesis (\(H_{0}\)) and the alternative hypothesis (\(H_{a}\)). The former, the null hypothesis, is a statement of no effect or no difference – in this case, stating that the true population proportion is 0.6. The latter, the alternative hypothesis, contradicts this by suggesting there is an effect or a difference – and in our problem, it proposes that the population proportion is greater than 0.6.

Decisions in hypothesis testing revolve around these two hypotheses. We evaluate the strength of evidence against the null hypothesis by calculating the p-value. If the evidence is strong (typically, p-value < 0.05), we may reject the null hypothesis in favor of the alternative hypothesis. Otherwise, we do not have sufficient evidence to reject the null hypothesis, and it stands.

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Most popular questions from this chapter

Influencing Voters Exercise 4.38 on page 235 describes a possible study to see if there is evidence that a recorded phone call is more effective than a mailed flyer in getting voters to support a certain candidate. The study assumes a significance level of \(\alpha=0.05\) (a) What is the conclusion in the context of this study if the p-value for the test is \(0.027 ?\) (b) In the conclusion in part (a), which type of error are we possibly making: Type I or Type II? Describe what that type of error means in this situation. (c) What is the conclusion if the p-value for the test is \(0.18 ?\) (d) In the conclusion in part (c), which type of error are we possibly making: Type I or Type II? Describe what that type of error means in this situation.

State the null and alternative hvpotheses for the statistical test described. Testing to see if there is evidence that a mean is less than 50

Exercises 4.117 to 4.122 give null and alternative hypotheses for a population proportion, as well as sample results. Use StatKey or other technology to generate a randomization distribution and calculate a p-value. StatKey tip: Use "Test for a Single Proportion" and then "Edit Data" to enter the sample information. Hypotheses: \(H_{0}: p=0.5\) vs \(H_{a}: p>0.5\) Sample data: \(\hat{p}=30 / 50=0.60\) with \(n=50\)

Mercury Levels in Fish Figure 4.26 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 68 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 pH causes the average mercury level in fish to increase? Why or why not?

In Exercises 4.150 to \(4.152,\) a confidence interval for a sample is given, followed by several hypotheses to test using that sample. In each case, use the confidence interval to give a conclusion of the test (if possible) and also state the significance level you are using. A 99\% confidence interval for \(\mu: 134\) to 161 (a) \(H_{0}: \mu=100\) vs \(H_{a}: \mu \neq 100\) (b) \(H_{0}: \mu=150\) vs \(H_{a}: \mu \neq 150\) (c) \(H_{0}: \mu=200\) vs \(H_{a}: \mu \neq 200\)

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