Consider a population proportion p = 0.78.
a-1. Calculate the expected value and the standard error of P−
with n = 40. (Round "expected value" to 2 decimal places and "standard deviation" to 4 decimal places.)
a-2. Is it appropriate to use the normal distribution approximation for P−
?
multiple choice 1
Yes
No
b-1. Calculate the expected value and the standard error of P−
with n = 40. (Round "expected value" to 2 decimal places and "standard deviation" to 4 decimal places.)
b-2. Is it appropriate to use the normal distribution approximation for P−
?
multiple choice 2
Yes
No
a-1. To calculate the expected value and standard error of P−, we need to use the formula:
Expected value of P− = p = 0.78
Standard error of P− = sqrt(pq/n) where q = 1 - p
Standard error of P− = sqrt(0.78 * (1 - 0.78)/40) ≈ 0.0513 (rounded to 4 decimal places)
Therefore, the expected value of P− is 0.78 and the standard error is 0.0513.
a-2. No, it is not appropriate to use the normal distribution approximation for P− when n = 40. The normal approximation is generally considered appropriate when both np and n(1-p) are greater than or equal to 5, where n is the sample size. In this case, np = 31.2 (40 * 0.78) and n(1-p) = 8.8 (40 * 0.22), which are both less than 5. Therefore, the normal distribution approximation is not appropriate.
b-1. The expected value and standard error of P− with n = 40 are the same as in part a-1:
Expected value of P− = p = 0.78
Standard error of P− = sqrt(pq/n) where q = 1 - p
Standard error of P− = sqrt(0.78 * (1 - 0.78)/40) ≈ 0.0513 (rounded to 4 decimal places)
Therefore, the expected value of P− is 0.78 and the standard error is 0.0513.
b-2. No, it is not appropriate to use the normal distribution approximation for P− when n = 40. The normal approximation is generally considered appropriate when both np and n(1-p) are greater than or equal to 5, where n is the sample size. In this case, np = 31.2 (40 * 0.78) and n(1-p) = 8.8 (40 * 0.22), which are both less than 5. Therefore, the normal distribution approximation is not appropriate.
a-1. To calculate the expected value and standard error of P− with n = 40, we can use the formulas:
Expected value (μ): μ = p = 0.78
Standard error (SE): SE = √[p(1-p)/n]
SE = √[(0.78)(1-0.78)/40]
SE ≈ 0.0688 (rounded to 4 decimal places)
Therefore, the expected value is 0.78 and the standard error is approximately 0.0688.
a-2. To determine if it is appropriate to use the normal distribution approximation for P−, we need to check if np and n(1-p) are both greater than 5. In this case, n = 40 and p = 0.78.
np = 40 * 0.78 = 31.2 (greater than 5)
n(1-p) = 40 * (1-0.78) = 8.8 (greater than 5)
Since both np and n(1-p) are greater than 5, it is appropriate to use the normal distribution approximation for P−.
a-2. The answer is Yes.
b-1. The expected value and standard error of P− with n = 40 will remain the same as in the previous question:
Expected value (μ): μ = p = 0.78
Standard error (SE): SE = √[p(1-p)/n]
SE = √[(0.78)(1-0.78)/40]
SE ≈ 0.0688 (rounded to 4 decimal places)
Therefore, the expected value is 0.78 and the standard error is approximately 0.0688.
b-2. To determine if it is appropriate to use the normal distribution approximation for P−, we need to check if np and n(1-p) are both greater than 5. In this case, n = 40 and p = 0.78.
np = 40 * 0.78 = 31.2 (greater than 5)
n(1-p) = 40 * (1-0.78) = 8.8 (greater than 5)
Since both np and n(1-p) are greater than 5, it is appropriate to use the normal distribution approximation for P−.
b-2. The answer is Yes.
a-1. To calculate the expected value and standard error of P̂ with n = 40, we can use the following formulas:
Expected value (mean) of P̂ = p
Standard error of P̂ = sqrt((p*(1-p))/n)
Given that p = 0.78 and n = 40, we can substitute these values into the formulas to find the expected value and standard error:
Expected value (mean) of P̂ = 0.78
Standard error of P̂ = sqrt((0.78*(1-0.78))/40)
Calculating the standard error using the formula:
Standard error of P̂ = sqrt((0.78*(1-0.78))/40) ≈ 0.0566 (rounded to 4 decimal places)
Therefore, the expected value (mean) of P̂ is 0.78 and the standard error is approximately 0.0566.
a-2. To determine if it's appropriate to use the normal distribution approximation for P̂, we need to consider the conditions for using the normal approximation to the binomial distribution. The conditions are:
1. np ≥ 10
2. n(1 - p) ≥ 10
Let's check if these conditions are satisfied with n = 40 and p = 0.78:
np = 40 * 0.78 = 31.2
n(1 - p) = 40 * (1 - 0.78) = 8.8
Since both np and n(1 - p) are greater than or equal to 10, it is appropriate to use the normal distribution approximation for P̂.
Therefore, the answer to multiple choice 1 is Yes.
b-1. The expected value (mean) of P̂ with n = 40 will be the same as in part a-1, which is 0.78.
The standard error of P̂ with n = 40 will also be the same as in part a-1, which is approximately 0.0566.
b-2. Similarly to part a-2, we need to check the conditions for using the normal approximation to the binomial distribution:
np = 40 * 0.78 = 31.2
n(1 - p) = 40 * (1 - 0.78) = 8.8
Since both np and n(1 - p) are greater than or equal to 10, it is appropriate to use the normal distribution approximation for P̂.
Therefore, the answer to multiple choice 2 is Yes.