People end up tossing 12% of what they buy at the grocery store (Reader's Digest, March 2009). Assume this is the true population proportion and that you plan to take a sample survey of 540 grocery shoppers to further investigate their behavior.
a.Show the sampling distribution of ( ), the proportion of groceries thrown out by your sample respondents (to 4 decimals)?
b. What is the probability that your survey will provide a sample proportion within ±.03 of the population proportion (to 4 decimals)?
c. What is the probability that your survey will provide a sample proportion within ±.015 of the population proportion (to 4 decimals)?
a. Hmmm, let's talk about the sampling distribution of the proportion of groceries thrown out. Since we are assuming a true population proportion of 12%, we can use this information to calculate the mean and standard deviation of the sampling distribution. The mean of the sampling distribution is going to be the same as the population proportion, which is 0.12. As for the standard deviation, we can calculate it using the formula sqrt((p(1-p))/n), where p is the population proportion and n is the sample size. Plugging in the numbers, we get sqrt((0.12*(1-0.12))/540), which gives us the standard deviation of the sampling distribution.
b. Now, let's move on to the probability that our survey will provide a sample proportion within ±0.03 of the population proportion. We need to find the z-score for ±0.03, using the formula z = (x - μ) / σ. In this case, x is ±0.03, μ is the population proportion of 0.12, and σ is the standard deviation we calculated earlier. Once you have the z-score, you can then look up the corresponding probability in the standard normal distribution table.
c. Finally, let's calculate the probability that our survey will provide a sample proportion within ±0.015 of the population proportion. Similar to the previous step, we need to find the z-score for ±0.015 and then look up the corresponding probability in the standard normal distribution table.
Wow, all these calculations just made me hungry! Perhaps I should take a small break and grab a snack from the grocery store... or maybe not, since 12% of it would just go to waste according to Reader's Digest!
To answer these questions, we need to use the sampling distribution of the proportion, also known as the binomial distribution. The parameters for this distribution are the population proportion (p) and the sample size (n).
Given:
- Population proportion (p) = 0.12
- Sample size (n) = 540
a. To find the sampling distribution of the proportion, we need to calculate the mean (μ) and standard deviation (σ) of the distribution.
The mean (μ) is given by:
μ = p
= 0.12
The standard deviation (σ) is given by:
σ = sqrt((p * (1 - p)) / n)
= sqrt((0.12 * (1 - 0.12)) / 540)
≈ sqrt(0.00020696)
≈ 0.0144 (rounded to 4 decimals)
Therefore, the sampling distribution of the proportion has a mean of 0.12 and a standard deviation of 0.0144.
b. To find the probability that the survey will provide a sample proportion within ±0.03 of the population proportion, we need to calculate the z-scores for both boundaries.
For the lower boundary:
z_lower = (0.12 - 0.03 - 0.12) / 0.0144
= -0.03 / 0.0144
≈ -2.0833 (rounded to 4 decimals)
For the upper boundary:
z_upper = (0.12 + 0.03 - 0.12) / 0.0144
= 0.03 / 0.0144
≈ 2.0833 (rounded to 4 decimals)
Next, we can use a standard normal distribution table or calculator to find the probabilities associated with these z-scores.
P(-2.0833 < Z < 2.0833) ≈ 0.955 (rounded to 4 decimals)
Therefore, the probability that the survey will provide a sample proportion within ±0.03 of the population proportion is approximately 0.955.
c. To find the probability that the survey will provide a sample proportion within ±0.015 of the population proportion, we follow the same steps as in part b.
For the lower boundary:
z_lower = (0.12 - 0.015 - 0.12) / 0.0144
= -0.015 / 0.0144
≈ -1.0417 (rounded to 4 decimals)
For the upper boundary:
z_upper = (0.12 + 0.015 - 0.12) / 0.0144
= 0.015 / 0.0144
≈ 1.0417 (rounded to 4 decimals)
Using a standard normal distribution table or calculator:
P(-1.0417 < Z < 1.0417) ≈ 0.706 (rounded to 3 decimals)
Therefore, the probability that the survey will provide a sample proportion within ±0.015 of the population proportion is approximately 0.706.
To answer these questions, we need to use the concept of sampling distribution and the Central Limit Theorem. Let's break down each question and explain how to calculate the answers.
a. Show the sampling distribution of ( ), the proportion of groceries thrown out by your sample respondents (to 4 decimals)?
To show the sampling distribution, we need to calculate the mean and standard deviation of the sample proportion. The mean of the sampling distribution is the same as the population proportion, which is 12% or 0.12.
The standard deviation of the sampling distribution (also known as the standard error) can be calculated using the formula:
Standard Deviation = sqrt( ( p * (1 - p) ) / n )
Where:
- p is the population proportion (0.12 in this case)
- n is the sample size (540 in this case)
So, let's plug in the values and calculate the standard deviation:
Standard Deviation = sqrt( ( 0.12 * (1 - 0.12) ) / 540 )
After calculating, the standard deviation of the sampling distribution turns out to be approximately 0.0117 (rounded to 4 decimals).
b. What is the probability that your survey will provide a sample proportion within ±.03 of the population proportion (to 4 decimals)?
To calculate this probability, we need to find the area under the sampling distribution curve within ±0.03. Since the sampling distribution follows a normal distribution, we can use Z-scores and the Standard Normal Table (Z-table).
First, we need to convert the ±0.03 to a Z-score. The formula to calculate the Z-score is:
Z = (x - μ) / σ
Where:
- x is the value we want to find the probability for (±0.03 in this case)
- μ is the mean of the sampling distribution (0.12 in this case)
- σ is the standard deviation of the sampling distribution (0.0117 in this case)
For the positive side, the Z-score is:
Z = (0.03 - 0.12) / 0.0117
After calculating, the positive Z-score is approximately -8.12 (rounded to 2 decimal places).
Next, we need to find the probability corresponding to this Z-score from the Standard Normal Table (Z-table). The Z-table provides the area under the curve up to a given Z-score.
Since we want the area within ±0.03, we need to find the area from -0.03 to +0.03. However, the Z-table only provides the area to the left of a Z-score. To find the area between ±0.03, we subtract the area to the left of -0.03 from the area to the left of +0.03.
Using the Z-table, the area to the left of -8.12 is practically zero, and the area to the left of +0.03 is approximately 0.509 (rounded to 3 decimal places).
So, the area between ±0.03 is:
Area = 0.509 - 0 = 0.509
Hence, the probability that the survey will provide a sample proportion within ±0.03 of the population proportion is approximately 0.509 (rounded to 4 decimals).
c. What is the probability that your survey will provide a sample proportion within ±0.015 of the population proportion (to 4 decimals)?
To calculate this probability, we follow a similar process as in part b. We need to calculate the Z-score and find the corresponding area under the sampling distribution curve.
For the positive side, the Z-score is:
Z = (0.015 - 0.12) / 0.0117
After calculating, the positive Z-score is approximately -8.12 (rounded to 2 decimal places).
Using the Z-table, the area to the left of -8.12 is practically zero, and the area to the left of +0.015 is approximately 0.373 (rounded to 3 decimal places).
So, the area between ±0.015 is:
Area = 0.373 - 0 = 0.373
Hence, the probability that the survey will provide a sample proportion within ±0.015 of the population proportion is approximately 0.373 (rounded to 4 decimals).