One of the ways most people relieve stress is to reward themselves with sweets. According
to one study in Food Technology, 46 percent admit to overeating sweet foods when
stressed. Suppose that the 46 percent figure is correct and that a random sample of 100
people is selected.
A. Does the distribution of, the sample proportion of people who relieve stress by
overeating sweet foods, have an approximately normal distribution? If so, what are its
mean and standard deviation?
B. What is the probability that the sample proportion, , exceeds 0.5?
C. What is the probability that lies within the interval 0.35 to 0.55?
D. What might you conclude if the sample proportion were as small as 30 percent?
To answer these questions, we will use the properties of the sampling distribution of a sample proportion.
A. The distribution of the sample proportion of people who relieve stress by overeating sweet foods will have an approximately normal distribution if certain conditions are met. One of these conditions is that the sample size is sufficiently large (n >= 30) and the population from which the sample is drawn is at least 10 times larger than the sample size.
In this case, the sample size is 100 people, which is larger than 30. However, we don't have information about the size of the population. So, we cannot definitively conclude whether the distribution of the sample proportion is approximately normal.
If the conditions are met, then the mean of the sample proportion is equal to the population proportion, which is 0.46 in this case. The standard deviation of the sample proportion can be calculated using the formula:
Standard deviation (σ) = sqrt((p * (1 - p)) / n)
where p is the population proportion and n is the sample size. Plugging in the values:
σ = sqrt((0.46 * (1 - 0.46)) / 100) = 0.0496
B. To find the probability that the sample proportion exceeds 0.5, we need to calculate the z-score and use the standard normal distribution table or a calculator.
z = (x - μ) / σ
where x is the value we want to find the probability for, μ is the mean of the sample proportion, and σ is the standard deviation.
In this case, x = 0.5, μ = 0.46, and σ = 0.0496. Plugging in the values:
z = (0.5 - 0.46) / 0.0496 = 0.08
Now, you can look up the probability corresponding to the z-score of 0.08 in the standard normal distribution table or use a calculator with a normal distribution function to find the probability.
C. To find the probability that the sample proportion lies within the interval 0.35 to 0.55, we can calculate the z-scores for both boundaries and find the difference between their probabilities.
For 0.35:
z1 = (x1 - μ) / σ
z1 = (0.35 - 0.46) / 0.0496
For 0.55:
z2 = (x2 - μ) / σ
z2 = (0.55 - 0.46) / 0.0496
Using the z-scores, you can find the probabilities associated with them in the standard normal distribution table or use a calculator to calculate the difference between the probabilities.
D. If the sample proportion were as small as 30%, we might conclude that a smaller proportion of people relieve stress by overeating sweet foods than we initially thought. However, we cannot make definitive conclusions based on a single sample. To be more confident in our conclusion, we would need to conduct further analysis and collect more data.