MATH1127: Introduction to Statistics (60022) Lesson 7.7 Using the Central Limit Theorem (Combination)
Unit 3 Chapter 7 : Les
Unit 3 Chapter 7: Lesson 7.7 Assignment
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Question 11
A Food Marketing Institute found that $28 \%$ of households spend more than $\$ 125$ a week on groceries. Assume the population proportion is 0.28 and a simple random sample of 256 households is selected from the population. What is the probability that the sample proportion of households spending more than $\$ 125$ a week is less than 0.25 ?
There is a probability that the sample proportion of households spending more than $\$ 125$ a week is less than 0.25 . Round the answer to 4 decimal places.
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Step 1 ：The problem is asking for the probability that the sample proportion of households spending more than $125 a week is less than 0.25. This is a problem of sampling distribution of proportions. We can use the Central Limit Theorem to solve this problem. The Central Limit Theorem states that if you have a population with mean μ and standard deviation σ and take sufficiently large random samples from the population with replacement, then the distribution of the sample means will be approximately normally distributed. This will hold true regardless of whether the source population is normal or skewed, provided the sample size is sufficiently large (usually n > 30).

Step 2 ：In this case, we know the population proportion (p) is 0.28, the sample size (n) is 256, and we are looking for the probability that the sample proportion (p̂) is less than 0.25.

Step 3 ：We can calculate the mean of the sampling distribution (μ_p̂) as p, and the standard deviation (σ_p̂) as \(\sqrt{p(1-p)/n}\).

Step 4 ：Then we can standardize the sample proportion to a z-score using the formula z = (p̂ - μ_p̂) / σ_p̂, and look up this z-score in the standard normal distribution to find the probability.

Step 5 ：Substituting the given values, we get p = 0.28, n = 256, p_hat = 0.25, mu_p_hat = 0.28, sigma_p_hat = \(\sqrt{0.28(1-0.28)/256}\) = 0.02806243040080456, z = (0.25 - 0.28) / 0.02806243040080456 = -1.0690449676496985.

Step 6 ：Looking up this z-score in the standard normal distribution, we find the probability to be 0.1425.

Step 7 ：Final Answer: The probability that the sample proportion of households spending more than $125 a week is less than 0.25 is \(\boxed{0.1425}\).