Question

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 born 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

Yes, the data suggest that the true proportion who remove the implants significantly differs for those that were born deaf and those that went deaf after learning to speak, at a .01 significance level.

Step-by-step solution

Define the samples and events

First, the two samples and the associated events need to be defined.The first sample consists of 250 adults born deaf, and the second sample consists of 250 adults who went deaf after learning to speak. The event of interest is the removal of the auditory nerve implant.

Formulate the null and alternative hypothesis

Based on the question, the null hypothesis \(H_0\) is that the proportions of implant removals among those deaf from birth \(p_1\) and those who went deaf after learning to speak \(p_2\) are the same. Thus, \(H_0: p_1 = p_2\). The alternative hypothesis \(H_a\) is that the proportions are not equal, i.e. \(H_a: p_1 ≠ p_2\). We are testing these hypotheses at a .01 significance level.

Calculate Sample Proportions

The proportions of adults who removed the implant should be calculated for each group. For those born deaf, this is 75 out of 250 or \(p_1 = 0.3\). For those who went deaf after learning to speak, this is 25 out of 250, or \(p_2 = 0.1\). Now, calculate the pooled proportion \(p\), which is the total number of successes divided by the total sample size, which equals (75+25)/(250+250) = 0.2.

Conduct a hypothesis test

Now, using a z-test for comparing two proportions, calculate the test statistic z. The z-score can be calculated using formula \[z = \frac{(p_1 - p_2) - 0}{\sqrt{p(1-p)\left(\frac{1}{n_1} + \frac{1}{n_2}\right) }}.\]With the given values, the z-score calculation gives approximately 6.1237 simply by substituting the values \(p_1 = 0.3\), \(p_2 = 0.1\), \(p = 0.2\), \(n_1 = n_2 = 250\).

Make a decision and interpret the result

The critical value for a two-tailed test at the .01 significance level is approximately ±2.58 (using a standard z-distribution table).As 6.1237 > 2.58, we reject the null hypothesis. There is significant evidence at the .01 level to 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.

Key concepts

Statistical Significance

Statistical significance plays a crucial role in hypothesis testing as it helps us determine whether the results of a study are likely due to chance or to some significant factor. In our scenario, we are comparing the proportion of implant removals between two groups of individuals with the aim of figuring out if the difference observed is statistically significant.

For this purpose, a significance level, generally denoted by \( \alpha \) is predetermined, which reflects the probability of rejecting the null hypothesis when it is actually true, known as a Type I error. For our study, we have chosen an \( \alpha \) of 0.01, representing only a 1% chance we are willing to accept for making this error.

If our test statistic — calculated based on sample data — falls beyond the critical value that corresponds to this \( \alpha \) in a standard distribution, we then conclude that our results are statistically significant and that the null hypothesis can be rejected. This means we're confident, with a 99% assurance, that the difference in implant removals between the two studied groups isn't due to random variation alone.

Proportions Comparison

The comparison of proportions is a statistical technique used to assess whether the proportion of occurrences of a certain event differs significantly between two groups. In the textbook exercise, we are examining whether the proportions of individuals who remove an auditory implant differ between those born deaf and those who became deaf later in life.

To compare proportions, we use the respective ratios from both samples: \( p_1 \) for those born deaf and \( p_2 \) for those who became deaf later. We compute these from our available data, finding \( p_1 = 0.3 \) and \( p_2 = 0.1 \) in our example.

However, just noting that \( p_1 \) is greater than \( p_2 \) is not enough. We also must determine if this difference in proportions is a random occurrence or a result of a genuine disparity. This leads us to perform a test of hypothesis where the null hypothesis states there is no difference in the proportions (\( H_0: p_1 = p_2 \) ) and the alternative suggests a difference exists (\( H_a: p_1 eq p_2 \)).

Z-test

A z-test is a type of statistical test used when conditions are suitable for approximating the distribution of sample proportions to a normal distribution. This is applicable when comparing two proportions as we have large sample sizes and the sample proportions are not too close to 0 or 1.

In the exercise, we apply a z-test to determine if the difference in implant removal proportions between the two groups is statistically significant. It involves calculating the z-score, a representation of how many standard deviations a data point (in this case, the difference between our sample proportions) is from the mean.

The formula for the z-score in the context of comparing two proportions is given by \[z = \frac{(p_1 - p_2) - 0}{\sqrt{p(1-p)(\frac{1}{n_1} + \frac{1}{n_2}) }}.\] After calculating our z-score to be approximately 6.1237, which is greater than the critical value of ±2.58, we can conclude significant evidence against the null hypothesis. Thus, the z-test provides a statistical backing to confidently state that there is a difference in the proportion of implant removals between the two groups studied.