Question

Are you a chocolate "purist," or do you like other ingredients in your chocolate? American Demographics reports that almost \(75 \%\) of consumers like traditional ingredients such as nuts or caramel in their chocolate. They are less enthusiastic about the taste of mint or coffee that provide more distinctive flavors. \({ }^{14}\) A random sample of 200 consumers is selected and the number who like nuts or caramel in their chocolate is recorded. a. What is the approximate sampling distribution for the sample proportion \(\hat{p}\) ? What are the mean and standard deviation for this distribution? b. What is the probability that the sample percentage is greater than \(80 \% ?\) c. Within what limits would you expect the sample proportion to lie about \(95 \%\) of the time?

Short answer

Answer: The probability is approximately 5.13%.

Step-by-step solution

Find the mean and standard deviation of the sample proportion distribution

The mean (\(\mu_{\hat{p}}\)) of the sample proportion is equal to the population proportion \(p\), so we have: \(\mu_{\hat{p}} = p = 0.75\) The standard deviation (\(\sigma_{\hat{p}}\)) of the sample proportion is given by the formula: \(\sigma_{\hat{p}} = \sqrt{\frac{p(1-p)}{n}}\) where \(n\) is the sample size. Plug in the values: \(\sigma_{\hat{p}} = \sqrt{\frac{0.75(1-0.75)}{200}} = \sqrt{\frac{0.75(0.25)}{200}} = \sqrt{0.0009375} = 0.0306\) So the mean of the sample proportion is 0.75, and the standard deviation is 0.0306.

Calculate the probability that the sample percentage is greater than 80%

To find the probability that the sample percentage is greater than 80%, we first convert the percentage to a proportion: \(\frac{80}{100} = 0.8\) Now we will use the z-score formula to find the z-value: \(z = \frac{\hat{p} - \mu_{\hat{p}}}{\sigma_{\hat{p}}} = \frac{0.8 - 0.75}{0.0306} = 1.634\) Using the standard normal distribution table or calculator, we find the area to the right of the z-score (greater than 1.634): \(P(Z > 1.634) = 1 - P(Z \leq 1.634) = 1 - 0.9487 = 0.0513\) So the probability that the sample percentage is greater than 80% is approximately 0.0513 or 5.13%.

Find the confidence interval for the sample proportion

To find the interval within which the sample proportion would lie about 95% of the time, we will use the formula for constructing a confidence interval: \(\hat{p} \pm z_{\alpha/2} \cdot \sigma_{\hat{p}}\) For a 95% confidence interval, \(z_{\alpha/2} = 1.96\). Now let's plug in the values we found earlier: \(0.75 \pm 1.96 \cdot 0.0306 = (0.75 - 0.059976, 0.75 + 0.059976) = (0.690024, 0.809976)\) Thus, we would expect the sample proportion to lie within the interval (0.690024, 0.809976) about 95% of the time.

Key concepts

Sample Proportion

The sample proportion, denoted as \( \hat{p} \), is simply the ratio of individuals in the sample who possess a certain characteristic to the total sample size. In our case, it represents the proportion of sampled consumers who prefer traditional ingredients like nuts or caramel in their chocolate. When we draw a sample from a larger population, we expect our sample proportion \( \hat{p} \) to be close to the true population proportion \( p \). This proximity forms the basis for making inferences about the population.
For our chocolate-loving consumers, if the sample size \( n \) is 200 and the number of those who prefer nuts and caramel is recorded, \( \hat{p} \) will give us an estimate of the population proportion. Regular sampling ensures that our results are representative of larger preferences among chocolate consumers.

Confidence Interval

The confidence interval provides a range of values within which we expect the true population proportion to fall, given a certain confidence level (e.g., 95%). It accounts for the sampling error, which can occur simply because not all observations of a population are included in the sample.
The confidence interval for a sample proportion is calculated using the formula \( \hat{p} \pm z_{\alpha/2} \cdot \sigma_{\hat{p}} \). Here, \( z_{\alpha/2} \) represents the z-score corresponding to the desired confidence level, while \( \sigma_{\hat{p}} \) is the standard deviation of the sample proportion. For a 95% confidence level, the formula includes a \( z \) value of 1.96. In the given exercise, the interval was calculated to be (0.690024, 0.809976), meaning we are 95% confident the true proportion of consumers who prefer nuts or caramel lies in this interval.

Standard Deviation

The standard deviation of the sample proportion \( \sigma_{\hat{p}} \) measures the amount of variability or dispersion of \( \hat{p} \) around the true population proportion. A smaller standard deviation indicates that \( \hat{p} \) is likely closer to \( p \).
The calculation formula is \( \sigma_{\hat{p}} = \sqrt{\frac{p(1-p)}{n}} \). Here, \( p \) is the population proportion, and \( n \) is the sample size. In the example, using a population proportion of 0.75 with a sample size of 200, the standard deviation was found to be approximately 0.0306. This value helps in constructing the confidence interval and in other inferential statistics.

Z-score

The Z-score is a statistical measurement that describes a value's relation to the mean of a group of values, expressed in terms of standard deviations. It is crucial in determining probabilities associated with the normal distribution.
For this exercise, the Z-score helps to calculate the likelihood that the sample proportion exceeds a specific value, like 80%. The formula used is \( z = \frac{\hat{p} - \mu_{\hat{p}}}{\sigma_{\hat{p}}} \), where \( \mu_{\hat{p}} \) is the mean of the sample proportion, which equals \( p \).
In the problem, a Z-score of 1.634 was found when checking if more than 80% of the sample prefer nuts or caramel. This indicates how many standard deviations \( \hat{p} \) is from the mean, allowing us to find probabilities using standard normal distribution tables or calculators.