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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

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.

Of all freshman at a large college, \(16 \%\) made the dean's list in the current year. As part of a class project, students randomly sample 40 students and check if those students made the list. They repeat this 1,000 times and build a distribution of sample proportions. (a) What is this distribution called? (b) Would you expect the shape of this distribution to be symmetric, right skewed, or left skewed? Explain your reasoning. (c) Calculate the variability of this distribution. (d) What is the formal name of the value you computed in (c)? (e) Suppose the students decide to sample again, this time collecting 90 students per sample, and they again collect 1,000 samples. They build a new distribution of sample proportions. How will the variability of this new distribution compare to the variability of the distribution when each sample contained 40 observations?

A study suggests that \(60 \%\) of college student spend 10 or more hours per week communicating with others online. You believe that this is incorrect and decide to collect your own sample for a hypothesis test. You randomly sample 160 students from your dorm and find that \(70 \%\) spent 10 or more hours a week communicating with others online. A friend of yours, who offers to help you with the hypothesis test, comes up with the following set of hypotheses. Indicate any errors you see. $$ \begin{array}{l} H_{0}: \hat{p}<0.6 \\ H_{A}: \hat{p}>0.7 \end{array} $$

Exercise 5.11 provides a \(95 \%\) confidence interval for the mean waiting time at an emergency room (ER) of (128 minutes, 147 minutes). Answer the following questions based on this interval. (a) A local newspaper claims that the average waiting time at this ER exceeds 3 hours. Is this claim supported by the confidence interval? Explain your reasoning. (b) The Dean of Medicine at this hospital claims the average wait time is 2.2 hours. Is this claim supported by the confidence interval? Explain your reasoning. (c) Without actually calculating the interval, determine if the claim of the Dean from part (b) would be supported based on a \(99 \%\) confidence interval?

Determine whether the following statement is true or false, and explain your reasoning: "With large sample sizes, even small differences between the null value and the observed point estimate can be statistically significant."

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