91Ó°ÊÓ

A nationwide study in 2003 indicated that about \(60 \%\) of college students with cell phones send and receive text messages with their phones. Suppose a simple random sample of \(n=1136\) college students with cell phones is obtained. (Source: promomagazine.com) (a) Describe the sampling distribution of \(\hat{p},\) the sample proportion of college students with cell phones who send or receive text messages with their phones. (b) What is the probability that 665 or fewer college students in the sample send and receive text messages with their cell phones? Is this result unusual? (c) What is the probability that 725 or more college students in the sample send and receive text messages with their cell phone? Is this result unusual?

Step by step solution

01

- Describe the sampling distribution of \(\backslash hat{p}\)

For a simple random sample of size \(n = 1136\), the sampling distribution of the sample proportion \(\hat{p}\) is approximately normal if the sample size is large enough. Check the conditions of the Central Limit Theorem: \(np \geq 10 \) and \(n(1-p) \geq 10\). Given \(p = 0.60\), calculate \(np = 1136 * 0.60 = 681.6\) and \(n(1-p) = 1136 * 0.40 = 454.4\). Both are greater than 10, so the distribution of \(\hat{p}\) can be approximated by a normal distribution with mean \( \mu_{\hat{p}} = p = 0.60\) and standard deviation \( \sigma_{\hat{p}} = \sqrt{\frac{p(1-p)}{n}} = \sqrt{\frac{0.60 * 0.40}{1136}} \approx 0.0145\). Thus, \(\hat{p}\) is normally distributed with mean 0.60 and standard deviation 0.0145.

02

- Calculate the probability for 665 or fewer students (Part b)

First, convert 665 students to the sample proportion: \hat{p} = \frac{665}{1136} = 0.5859\. To find the probability that 665 or fewer students send and receive text messages, convert the sample proportion to the z-score: \(z = \frac{\hat{p} - \mu_{\hat{p}}}{\sigma_{\hat{p}}} = \frac{0.5859 - 0.60}{0.0145} = -0.9759\). Using the z-table, find the corresponding probability: \( P(Z \leq -0.9759) \approx 0.164 \) or 16.4%. Since this probability is greater than 5%, the result is not considered unusual.

03

- Calculate the probability for 725 or more students (Part c)

Convert 725 students to the sample proportion: \hat{p} = \frac{725}{1136} = 0.638\. Convert this sample proportion to the z-score: \(z = \frac{0.638 - 0.60}{0.0145} = 2.62\). Use the z-table to find the corresponding probability: \( P(Z \geq 2.62) \approx 0.0044 \) or 0.44%. Since this probability is less than 5%, the result is considered unusual.

04

- Summarize the results for the probabilities

From the calculations, the probability that 665 or fewer students send and receive text messages is about 16.4% (not unusual), while the probability that 725 or more students do so is about 0.44% (unusual).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Central Limit Theorem

The Central Limit Theorem (CLT) is essential in statistics. It tells us that the sampling distribution of the sample mean (or sample proportion) will be approximately normal, no matter the shape of the population distribution, provided the sample size is large enough.
For proportions, if you have a large enough sample size, conditions to check include:

• (np) should be at least 10
• (n(1-p)) should be at least 10

Consider the exercise above, where the population proportion p is 0.60 and sample size n is 1136. Calculations give us np = 1136 * 0.60 = 681.6 and n(1-p) = 1136 * 0.40 = 454.4. Both values are larger than 10, so the sampling distribution can be approximated by a normal distribution.

Normal Distribution

A normal distribution is a symmetric, bell-shaped distribution where most of the observations cluster around the central peak.
In the context of the exercise, we found that the sampling distribution of the sample proportion is normal. It has the following parameters:

• Mean (μ) = population proportion (p) = 0.60
• Standard Deviation (σ) = sqrt[(p(1-p))/n] = sqrt[(0.60*0.40)/1136] ≈ 0.0145

This normal distribution makes it easier to calculate probabilities and find z-scores.

Probability Calculation

The next step involves calculating probabilities using the normal distribution.
For instance, to find the probability that 665 or fewer students send and receive texts, we convert this number to a sample proportion, which gives us 0.5859 (665/1136 ≈ 0.5859).
We then use the normal distribution parameters to determine the z-score and find the probability using standard normal distribution tables.
Here’s how it’s done:

• Z = (Sample Proportion - Mean) / Standard Deviation
• For 665 or fewer students, the z-score = (0.5859 - 0.60) / 0.0145 ≈ -0.976
• Using a z-table, we get: P(Z ≤ -0.976) ≈ 0.164 or 16.4%

Similar calculations are done for finding probabilities of other events as well.

Z-score Transformation

Z-scores help us standardize different data points, allowing us to compare them within the same context. A z-score measures the number of standard deviations a data point is from the mean.
The formula for z-score is given by: li>(Sample Proportion - Mean) / Standard Deviation
In our exercise for 725 or more students, we first convert the number of students to the sample proportion (0.638). Then, we calculate the z-score as follows:

• Z = (0.638 - 0.60) / 0.0145 ≈ 2.62
• Looking this up in the z-table, we find the probability: P(Z ≥ 2.62) ≈ 0.0044 or 0.44%.

This calculated z-score and corresponding probability allow us to determine whether certain results are unusual or not.