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. Well, if we're going to investigate people's grocery behavior, I hope they won't throw any rotten tomatoes at us! Now, let's calculate the sampling distribution of the proportion of groceries thrown out by our sample respondents.
Since we know the true population proportion is 12% or 0.12, and we're going to survey 540 grocery shoppers, we can use the standard deviation formula for proportions:
Standard deviation (σ) = sqrt[(p*q)/n]
Where p is the population proportion, q is 1-p, and n is the sample size.
σ = sqrt[(0.12*0.88)/540]
σ ≈ 0.0145
Now, we can use this standard deviation to calculate the sampling distribution using the central limit theorem. The sampling distribution will be normally distributed, with the mean equal to the population proportion and the standard deviation equal to the calculated standard deviation.
So, the sampling distribution of the proportion of groceries thrown out by our sample respondents is approximately N(0.12, 0.0145).
b. Now, let's calculate the probability that our survey will provide a sample proportion within ±0.03 of the population proportion.
To do this, we'll find the z-scores corresponding to ±0.03:
z1 = (0.03 - 0.12) / 0.0145
z2 = (-0.03 - 0.12) / 0.0145
Using a standard normal distribution table or a calculator, we can find the probabilities associated with these z-scores and then subtract them to find the probability between them.
P(0.09 < p < 0.15) = P(z1 < z < z2)
Now, I would love to give you the exact probability, but my crystal ball says it's out of stock. However, you can easily calculate this using your favorite statistical software or online calculator.
c. Similarly, let's calculate the probability that our survey will provide a sample proportion within ±0.015 of the population proportion.
To do this, we'll find the z-scores corresponding to ±0.015:
z1 = (0.015 - 0.12) / 0.0145
z2 = (-0.015 - 0.12) / 0.0145
Using the same method as in part b, we can find the probability between these z-scores.
P(0.105 < p < 0.135) = P(z1 < z < z2)
Again, you'll have to roll up your sleeves and calculate this probability using the appropriate statistical tools. Happy math-ing!
a. To show the sampling distribution of the proportion of groceries thrown out by the sample respondents, we use the formula for the standard error of a proportion:
Standard Error (SE) = sqrt[(p * q) / n]
where p is the population proportion, q is 1 - p, and n is the sample size.
Given:
- Population proportion (p) = 0.12
- Sample size (n) = 540
Plugging in the values:
SE = sqrt[(0.12 * (1 - 0.12)) / 540]
= sqrt[0.01296 / 540]
≈ 0.0146 (rounded to 4 decimal places)
Therefore, the standard error of the proportion is approximately 0.0146.
b. To calculate 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 the upper and lower bounds and then find the probability using the z-table.
Lower Bound:
z_lower = (sample proportion - population proportion) / SE
= (0.12 - 0.03) / 0.0146
≈ 6.164 (rounded to 4 decimal places)
Upper Bound:
z_upper = (sample proportion - population proportion) / SE
= (0.12 + 0.03) / 0.0146
≈ 9.589 (rounded to 4 decimal places)
Next, we find the probabilities associated with these z-scores using the z-table:
p_lower = P(z ≤ -6.164) ≈ 0 (rounded to 4 decimal places)
p_upper = P(z ≤ 9.589) ≈ 1 (rounded to 4 decimal places)
Now, the probability that the survey will provide a sample proportion within ±0.03 of the population proportion is:
p_within_0.03 = p_upper - p_lower
= 1 - 0
= 1
c. To calculate 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.
Lower Bound:
z_lower = (sample proportion - population proportion) / SE
= (0.12 - 0.015) / 0.0146
≈ 6.164 (rounded to 4 decimal places)
Upper Bound:
z_upper = (sample proportion - population proportion) / SE
= (0.12 + 0.015) / 0.0146
≈ 9.589 (rounded to 4 decimal places)
Next, we find the probabilities associated with these z-scores using the z-table:
p_lower = P(z ≤ -6.164) ≈ 0 (rounded to 4 decimal places)
p_upper = P(z ≤ 9.589) ≈ 1 (rounded to 4 decimal places)
Now, the probability that the survey will provide a sample proportion within ±0.015 of the population proportion is:
p_within_0.015 = p_upper - p_lower
= 1 - 0
= 1
a. To show the sampling distribution of the proportion of groceries thrown out by your sample respondents, you need to calculate the standard deviation of the sampling distribution. The formula for the standard deviation of a sampling distribution is:
standard deviation = sqrt((p * (1 - p)) / n)
where p is the population proportion (12% or 0.12) and n is the sample size (540).
Therefore, the standard deviation of the sampling distribution is:
standard deviation = sqrt((0.12 * (1 - 0.12)) / 540)
To find the sampling distribution to 4 decimals, we can use the sample mean (p-hat) as an estimate of the population proportion (p):
sampling distribution = 0.12 ± standard deviation
Therefore, the sampling distribution of the proportion of groceries thrown out by your sample respondents is approximately:
0.12 ± 0.0126
b. To find the probability that your survey will provide a sample proportion within ±0.03 of the population proportion, you need to find the z-score for a range of ±0.03 around the population proportion.
The formula to calculate the z-score is:
z-score = (sample proportion - population proportion) / standard deviation
For the range of ±0.03, the z-scores are:
(lower z-score) = (0.12 - 0.03) / 0.0126
(upper z-score) = (0.12 + 0.03) / 0.0126
Next, you can use a standard normal distribution table or a statistical software to find the corresponding probabilities for the z-scores. Subtract the probability corresponding to the lower z-score from the probability corresponding to the upper z-score to find the probability that your survey will provide a sample proportion within ±0.03 of the population proportion.
c. To find the probability that your survey will provide a sample proportion within ±0.015 of the population proportion, you can follow the same steps as in part b. Calculate the z-scores for a range of ±0.015 around the population proportion, and then find the corresponding probabilities for the z-scores using a standard normal distribution table or a statistical software. Subtract the probability corresponding to the lower z-score from the probability corresponding to the upper z-score to find the probability that your survey will provide a sample proportion within ±0.015 of the population proportion.