or Part A, we can use the formula for confidence intervals:
CI = p ± Zsqrt((pq)/n)
where p is the proportion of almonds found in the sample (13/100 = 0.13), q is 1-p, Z is the z-value for a 99% confidence level (2.576), and n is the sample size (100). Plugging in these values, we get:
CI = 0.13 ± 2.576sqrt((0.130.87)/100)
which simplifies to:
CI = (0.045, 0.215)
Therefore, we are 99% confident that the true proportion of almonds in the population of planters mixed nuts is between 0.045 and 0.215.
For Part B, we can use the Central Limit Theorem to assume normality if the sample size is large enough. Since n = 100 is greater than or equal to 30, we can assume normality.
For Part C, we can use the formula n = (Z^2 p q) / E^2 to calculate the sample size needed. Here, Z is the z-value for a 99% confidence level (2.576), p is the estimated proportion of almonds (0.13), q is 1-p, and E is the maximum error of the estimate in decimal form (0.05). Plugging in these values, we get:
n = (2.576^2 0.13 0.87) / 0.05^2
which simplifies to approximately 276. Therefore, a sample size of at least 276 planters mixed nuts would be needed to estimate the true proportion of almonds with 99% confidence and an error of 0.05