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An electronic implant that stimulates the auditory nerve has been used to restore partial hearing to a number of deaf people. In a study of implant acceptability (Los Angeles Times, January 29,1985 ), 250 adults bom deaf and 250 adults who went deaf after learning to speak were followed for a period of time after receiving an implant. Of those deaf from birth, 75 had removed the implant, whereas only 25 of those who went deaf after learning to speak had done so. Does this suggest that the true proportion who remove the implants differs for those that were born deaf and those that went deaf after learning to speak? Test the relevant hypotheses using a .01 significance level.

Short Answer

Expert verified
To conclude whether the true proportion who remove the implants differs for those that were born deaf and those that went deaf after learning to speak, we first state the null and alternative hypotheses. Then we calculate the sample proportions and test statistic. Following this, the p-value is computed and compared to the given significance level. The decision to reject or fail to reject the null hypothesis is based on this comparison.

Step by step solution

01

State the Null and Alternative Hypotheses

The null hypothesis (\(H_0\)) is that the true proportions removing implants for those born deaf and those who went deaf after learning to speak are equal. The alternative hypothesis (\(H_a\)) is that these proportions differ. This can be represented as: \(H_0: p_1 = p_2\) and \(H_a: p_1 ≠ p_2\) where \(p_1\) is the proportion removing implants for those born deaf and \(p_2\) is the proportion for those who went deaf after learning to speak.
02

Calculate Sample Proportions

For those born deaf, 75 out of 250 removed the implant, so \(\hat{p}_1 = 75/250 = 0.3\). For those who went deaf after learning to speak, 25 out of 250 removed the company implant, so \(\hat{p}_2 = 25/250 = 0.1\).
03

Calculate the Test Statistic

The formula for the test statistic for the difference in proportions is \(Z = \frac{(\hat{p}_1 - \hat{p}_2) - 0} {\sqrt{\hat{p}(1-\hat{p})(1/n_1 + 1/n_2)}}\), where \(\hat{p} = (x_1 + x_2) / (n_1 + n_2)\), hence, \(Z = \frac{(0.3 - 0.1) - 0} {\sqrt{0.2*(1-0.2)*(1/250 + 1/250)}}\)
04

Compute the p-value

The p-value is found by looking up the computed value of Z in a standard normal table. This is a two-tailed test, so the p-value is two times the area to the right of Z in the normal distribution.
05

Make a Decision

If the p-value is less than the significance level α=0.01, then reject the null cannot hypothesis. If the p-value is greater than α, 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.

Null Hypothesis
When conducting a hypothesis test, the first step is to clearly define the null hypothesis, often denoted as \(H_0\). In this context, the null hypothesis states that there is no difference between two population proportions. It essentially assumes that any observed difference is due to random chance.
For the exercise at hand, \(H_0\) posits that the proportion of people who remove the implants is the same for those born deaf and those who went deaf after learning to speak. Mathematically, this is represented as \(p_1 = p_2\), where \(p_1\) and \(p_2\) are the proportions for each respective group.
The null hypothesis acts as a default assumption, meaning that we maintain it until we have enough evidence to suggest otherwise. Additionally, rejecting the null hypothesis implies that an external factor may be affecting the observed data.
Alternative Hypothesis
The alternative hypothesis is the counterpart to the null hypothesis, and is symbolized as \(H_a\). It aims to show that there is a statistically significant effect or difference that the null hypothesis fails to predict.
In the given exercise, the alternative hypothesis suggests that the proportion of individuals who remove the implants is different between the two groups. This is mathematically presented as \(p_1 eq p_2\). The alternative hypothesis can be one-sided or two-sided. In this case, it is two-sided, reflecting that either group could have a higher proportion of removals.
The acceptance of the alternative hypothesis over the null hypothesis indicates meaningful differences tied to real-world phenomena, rather than randomness.
Significance Level
The significance level, commonly denoted as \(\alpha\), plays a pivotal role in hypothesis testing. It represents the probability of rejecting the null hypothesis when it is true—commonly referred to as a "Type I error."
For the exercise, \(\alpha\) is set at 0.01, indicating a 1% risk of incorrectly rejecting \(H_0\). The chosen level directly influences the test's rigor: a smaller \(\alpha\) implies greater caution in declaring statistical significance, as more compelling evidence is required to reject the null hypothesis.
Significance levels are typically set before conducting the test to prevent bias. In practice, conventional levels like 0.05 or 0.01 are used, balancing risk tolerance and test practicality.
P-value
In hypothesis testing, the p-value serves as a measure of the strength of the evidence against the null hypothesis. A calculated p-value quantifies the probability of observing the test result, or more extreme, given that the null hypothesis is true.
In our scenario, after computing the test statistic \(Z\), the next step is to find the p-value. By checking a standard normal distribution table, you can determine how likely it is to observe a \(Z\) value as extreme as the calculated one, under the assumption that \(H_0\) holds. A two-tailed test is applied here, doubling the p-value from one tail of the distribution.
Concluding the test depends on a comparison between the p-value and the pre-defined significance level \(\alpha\). If the p-value is lower than \(\alpha=0.01\), we reject the null hypothesis, suggesting that the observed difference is statistically significant and unlikely to occur by chance.

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