1. Sampling Distribution of Sample Mean
Use the "Grocery_shopping" data set to complete question 1. This data set recorded the total grocery spending for 50 consumers at a local grocery store.
a. Use software to find the descriptive statistics of the variable Spending. Copy and paste your output below and answer the following questions: If you need to review how to do this, see question 4 on lab activity 2!
i. What is the sample size?
ii. What is the mean?
iii. What is the standard deviation?
b. In many situations, such as this, we do not know the population standard deviation, so we use the sample standard deviation instead: ., also called the standard error. Calculate the standard error of the sample mean. Round your answer to two decimal places.
c. Now assume that the true average spending is µ = $28 and standard deviation is σ = $10. Using this information, follow these steps to finding P($25 < < $28) when n = 50.
i. What is the standard error of the sample mean? Round your answer to two decimal places.
ii. Fill in the blanks: The sample mean follows a ___________ distribution with mean ___________ and standard deviation _________.
Hint: Notice the sentence you are completing is about the sample mean. You should use the sample standard deviation value from part (i).
iii. Find the z-score for X = $25. Round your answer to two decimal places.
Hint: You can find the formula to calculate a z-score on page 282 of the textbook.
iv. Find the z-score for X = 28. Round your answer to two decimal places.
v. Recall that P(a < Z < b) = P(Z < b) - P(Z < a). Using this and the values you calculated in parts (iii and iv), find P($25 < < $28) using the Standard Normal Table.
vi. Interpret the value in part (v).
vii. Does the sample mean value in your dataset (from part (a)) seem unusual to you based on this information? Why or why not?
d. Still assuming that the true average spending is µ = $28 and standard deviation is σ = $10, find
P( ≤ $24) when n = 50. Round your answer to 4 decimal places.
Hint: First find the z-score. Note that you can use the standard error found in 1ci since the sample size and population standard deviation are the same. Then, use the Standard Normal Table to find the probability of that z-score, keeping in mind that the probability you get from the table is a ≤ probability.
e. Still assuming that the true average spending is µ = $28 and standard deviation is σ = $10, find P( > $24) when n = 50. Round your answer to 4 decimal places.
Hint: When you add your answers to part d and part e together, you should get 1 since part d is the < probability and part e is the > probability.
f. Now suppose that we take a sample of size n = 100 from the population described in part (c). Compared to when n = 50, would you expect the sample mean to be closer to or farther from the true population mean? Why?
2. Sampling Distribution of Sample Proportion
Suppose that at a blood drive a total of n = 300 participants donate blood, and the blood type for each donor is recorded. A "universal" donor is one who has a blood type of O negative. A blood drive set a goal of at least 40 universal donors during the event, but only 25 people donated O negative blood.
No data set needed.
a. What is the point estimate for the population proportion of universal donors, based on those who actually donated O negative blood? Recall that a point estimate for the population proportion is the sample proportion (, called p-hat).
b. According to the American Red Cross, the true proportion of universal donors in the U.S. population is p = 0.07.
i. Using the given, calculate standard error of the sample proportion, . Round your answer to four decimal places.
ii. Verify that normal approximation methods would be appropriate for estimating the distribution of the sample proportion (state the assumptions and check if they hold).
1. Write the assumptions that need to be met.
Hint: Review the online notes section "Sampling Distributions for Sample Proportion, p-hat" section for the assumptions.
2. Verify the assumptions are met in this case.
iii. Give the interval of values within which the sample proportion, 0.083, will almost certainly fall.
iv. Given that there were 300 participants in the blood drive, is it surprising that they did not meet their goal of at least 40 universal donors, a proportion of 0.13? Why or why not?