91Ó°ÊÓ

The sample proportion, denoted as \(\hat{p}\), represents the ratio of a specific characteristic within a sample to the total sample size. In this exercise, \(\hat{p}\) refers to the proportion of 'cell-phone only' users in a sample of 750 people. The formula to calculate \(\hat{p}\) is straightforward: divide the number of 'cell-phone only' users by the total number of users in the sample. For example, if 60 out of the 750 users rely only on cell phones, the sample proportion \(\hat{p}\) would be \left(\frac{60}{750}\right= 0.08\.

The principle of normal approximation comes into play when the sample size is sufficiently large. It allows us to approximate the sampling distribution of \(\hat{p}\) using a normal distribution. For this approximation to hold, two conditions must be met: \(np\) and \(n(1-p)\) should be greater than or equal to 10. In this exercise, \(n = 750\) and \(p = 0.07\). Therefore, \(np = 750 \cdot 0.07 = 52.5\) and \(n(1-p) = 750 \cdot 0.93 = 697.5\), both exceeding 10. As a result, we can say that the sampling distribution of \(\hat{p}\) is approximately normal with mean \(p = 0.07\) and standard deviation \[\sigma_{\hat{p}} = \sqrt{ \frac{ p(1-p) }{ n } } = 0.0095\].

The z-score is a measure that describes how many standard deviations a specific sample proportion \(\hat{p}\) is away from the population proportion \(p = 0.07\). The calculation of the z-score in this exercise is essential for determining the probability of an event. For instance, to find the probability that more than 8% of users are 'cell-phone only,' we convert 8% to a proportion \(\hat{p} = 0.08\) and compute the z-score as follows:
\[ Z = \frac{ \hat{p} - p }{ \sigma_{\hat{p}} } = \frac{ 0.08 - 0.07 }{ 0.0095 } = 1.05 \].
The z-score \(1.05\) indicates that the sample proportion is 1.05 standard deviations above the mean proportion.

Probability calculations help determine the likelihood of an event within a sampling distribution. Using the z-score calculated in the previous section, we look up the corresponding probability in a z-table. A z-score of \(1.05\) corresponds to a probability of approximately \(0.1477\) for the lower tail. Therefore, the probability of obtaining more than 8% 'cell-phone only' users is \(1 - 0.1477 = 0.8523\), or 85.23%. This high probability indicates that it's not unusual for more than 8% of users to rely solely on cell phones in a sample of 750 users.

The population proportion, denoted as \(p\), is the ratio of individuals in the entire population with a characteristic of interest. In the given exercise, the population proportion \(p = 0.07\) reflects that 7% of the population rely solely on cell phones.
This proportion is crucial for forming the sampling distribution and is used to determine if specific outcomes in the sample are unusual. For instance, having 40 or fewer 'cell-phone only' users in a sample of 750 represents a sample proportion of \(\hat{p} = 0.0533\). Calculating the z-score for \(\hat{p} = 0.0533\) gives:
\[ Z = \frac{ \hat{p} - p }{ \sigma_{\hat{p}} } = -1.76 \].
This corresponding probability, approximately \(0.0392\), is less than \(0.05\), indicating that such an event is unusual.