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Chapter 5: Problem 30

It is believed that nearsightedness affects about \(8 \%\) of all children. In a random sample of 194 children, 21 are nearsighted. Conduct a hypothesis test for the following question: do these data provide evidence that the \(8 \%\) value is inaccurate?

Short Answer

Expert verified
The data do not provide sufficient evidence to claim that the 8% value is inaccurate.

Step by step solution

01

Define Hypotheses

Start by defining the null and alternative hypotheses. The null hypothesis (H0) is that the proportion of nearsighted children is 0.08. The alternative hypothesis (H1) is that the proportion of nearsighted children is not 0.08. Formally, \( H_0: p = 0.08 \) and \( H_1: p eq 0.08 \).
02

Calculate the Sample Proportion

The sample proportion, \( \hat{p} \), is found by dividing the number of nearsighted children by the total number of children in the sample. Here, it is \( \hat{p} = \frac{21}{194} \approx 0.1082 \).
03

Determine Standard Error

The standard error (SE) of the sample proportion can be calculated using the formula \( SE = \sqrt{\frac{p(1-p)}{n}} \), where \( p = 0.08 \) and \( n = 194 \). Thus, \( SE = \sqrt{\frac{0.08 \times 0.92}{194}} \approx 0.0196 \).
04

Calculate the Z-statistic

Compute the Z-statistic using the formula \( Z = \frac{\hat{p} - p}{SE} \). Substituting in the values, \( Z = \frac{0.1082 - 0.08}{0.0196} \approx 1.439 \).
05

Find p-value

Use the Z-statistic to find the two-tailed p-value from the standard normal distribution. The Z-statistic of 1.439 corresponds to a p-value of approximately 0.1501 for a two-tailed test.
06

Make a Decision

Decide whether to reject or fail to reject the null hypothesis. If the p-value is greater than the significance level (often \(\alpha = 0.05\)), we fail to reject the null hypothesis. Here, since 0.1501 > 0.05, we fail 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.

Proportion Hypothesis Test
The hypothesis test for proportions helps us determine if a sample proportion is significantly different from a hypothesized population proportion. In this context, we are examining if the proportion of nearsighted children in a sample deviates from the believed 8% in the wider population.

We start by setting up two hypotheses:
  • Null Hypothesis (\( H_0 \)): The proportion of nearsighted children is 0.08.
  • Alternative Hypothesis (\( H_1 \)): The proportion of nearsighted children is not 0.08. This makes it a two-tailed test, as we are looking for any difference.
By comparing the sample proportion to the hypothesized proportion under these conditions, we can use statistical methods to determine if there's evidence suggesting the assumed proportion isn't accurate.
Standard Error Calculation
Calculating the standard error (SE) is crucial in understanding the variability expected in the sample proportion compared to the population proportion. The SE provides a measure of how much sampling variability we might expect. It is derived using the formula: \[ SE = \sqrt{\frac{p(1-p)}{n}} \] Here, \( p \) is the population proportion, \( 0.08 \), and \( n \) is the sample size, 194.

This calculates to about 0.0196, meaning the variability around the sample proportion that we expect due to random sampling is around this value. The smaller the SE, the more accurate our sample proportion is when estimating the true population proportion.
Z-statistic
The Z-statistic helps us understand how far the sample proportion is from the hypothesized population proportion, in terms of standard deviations. We calculate it with the formula:
  • \( Z = \frac{\hat{p} - p}{SE} \)
Let's break it down:
  • \( \hat{p} \) is the sample proportion, which is about 0.1082.
  • \( p \) is the population proportion, assumed to be 0.08.
  • SE is the calculated standard error, approximately 0.0196.
Substituting these values:
  • \( Z = \frac{0.1082 - 0.08}{0.0196} \approx 1.439 \).
This Z-statistic expresses the distance of the sample proportion from the hypothesized proportion in standard error units.
P-value Computation
The p-value helps us determine the statistical significance of our results. A p-value is the probability of observing a sample statistic as extreme as the one we obtained, assuming the null hypothesis is true.

Using our calculated Z-statistic of 1.439, we use statistical tables or software to find the p-value in a standard normal distribution. For a two-tailed test:
  • If Z equals 1.439, the p-value is around 0.1501.
This p-value indicates the likelihood of obtaining the observed data if the null hypothesis is correct. Since 0.1501 is greater than a typical threshold of 0.05, we do not reject the null hypothesis. Therefore, the data does not provide strong enough evidence to say the real proportion of nearsighted children differs from the supposed 8%.

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Most popular questions from this chapter

A study suggests that the \(25 \%\) of 25 year olds have gotten married. You believe that this is incorrect and decide to collect your own sample for a hypothesis test. From a random sample of 25 year olds in census data with size 776 , you find that \(24 \%\) of them are married. A friend of yours offers to help you with setting up the hypothesis test and comes up with the following hypotheses. Indicate any errors you see. $$ \begin{array}{l} H_{0}: \hat{p}=0.24 \\ H_{A}: \hat{p} \neq 0.24 \end{array} $$

As part of a quality control process for computer chips, an engineer at a factory randomly samples 212 chips during a week of production to test the current rate of chips with severe defects. She finds that 27 of the chips are defective. (a) What population is under consideration in the data set? (b) What parameter is being estimated? (c) What is the point estimate for the parameter? (d) What is the name of the statistic can we use to measure the uncertainty of the point estimate? (e) Compute the value from part (d) for this context. (f) The historical rate of defects is \(10 \%\). Should the engineer be surprised by the observed rate of defects during the current week? (g) Suppose the true population value was found to be \(10 \%\). If we use this proportion to recompute the value in part (e) using \(p=0.1\) instead of \(\hat{p},\) does the resulting value change much?

In each part below, there is a value of interest and two scenarios (I and II). For each part, report if the value of interest is larger under scenario I, scenario II, or whether the value is equal under the scenarios. (a) The standard error of \(\hat{p}\) when (I) \(n=125\) or (II) \(n=500\). (b) The margin of error of a confidence interval when the confidence level is (I) \(90 \%\) or (II) \(80 \%\). (c) The p-value for a Z-statistic of 2.5 calculated based on a (I) sample with \(n=500\) or based on a (II) sample with \(n=1000\). (d) The probability of making a Type 2 Error when the alternative hypothesis is true and the significance level is (I) 0.05 or (II) 0.10 .

In Exercise \(5.19,\) we learned that a Rasmussen Reports survey of \(1,000 \mathrm{US}\) adults found that \(42 \%\) believe raising the minimum wage will help the economy. Construct a \(99 \%\) confidence interval for the true proportion of US adults who believe this.

In 2013, the Pew Research Foundation reported that " \(45 \%\) of U.S. adults report that they live with one or more chronic conditions". \(^{12}\) However, this value was based on a sample, so it may not be a perfect estimate for the population parameter of interest on its own. The study reported a standard error of about \(1.2 \%\), and a normal model may reasonably be used in this setting. Create a \(95 \%\) confidence interval for the proportion of U.S. adults who live with one or more chronic conditions. Also interpret the confidence interval in the context of the study.
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