Determine the mean and standard deviation of the 500 sample means. Round to two decimal places: Hint: In StatCrunch, find the mean and standard deviation of the column “xbar: d) Are these values close to what was expected in part a) of Question 2? What proportion of the 500 random samples resulted in sample mean number of almonds greater than 53? Directions: Stat Tables Frequency Select Column(s): xbar; Where_ click “Build ; Columns: xbar; click “Add Column enter Statistic(s): Frequency Compute. Record the Count” for xbar f) Based on the normal model; what proportion of random samples of size n = 10 would we expect to result in : sample mean greater than 53? Hint: In StatCrunch; find the area under the standard normal curve using the mean and standard deviation from part with P(x 2 53). 9) Is the theoretical proportion obtained from the model (hint: refer to part – f) close to the proportion from the simulation? Explain:
The Correct Answer and Explanation is:
Answer:
To determine the mean and standard deviation of the 500 sample means (x̄), use StatCrunch:
- Find the mean and standard deviation:
- Go to Stat > Summary Stats > Columns
- Select the column xbar
- Click Compute
- Suppose the output gives a mean of 51.83 and a standard deviation of 1.45 (these values are examples and should match your actual output from StatCrunch).
- Are these values close to what was expected in part a of Question 2?
- In part a, you likely calculated the expected mean (μ) and standard deviation (σ/√n) using population values. Suppose the population mean was 52 and the population standard deviation was 4.6, then: Expected standard deviation of sample means=4.610≈1.46\text{Expected standard deviation of sample means} = \frac{4.6}{\sqrt{10}} \approx 1.46Expected standard deviation of sample means=10
