Suppose a simple random sample of size n=150 is obtained from a population whose size is N=20,000 and whose population proportion with a specified characteristic is p=0.8. Please answer questions (1) through (5) below.
(1) Describe the sampling distribution of p^ (choose the correct phrase that best describes the shape of the sampling below)
a) Not normal because n <, 0.05N and np (1-p) <10.
b) Approximately normal because n <, 0.05N and np (1-p) >10.
c) Approximately normal because n <, 0.05N and np (1-p) <10.
d) Not normal because n <, 0.05N and np (1-p) >10.
2) Determine the mean of the sampling distribution of p^.
p^ = ______ (round to one decimal place as needed)
3) Determine the standard deviation of the sampling distribution of p^.
p^ = ______ (round six decimal places as needed)
4) What is the probability of obtaining x=123 or more individuals with the characteristic? that is , what P(p^ > 0.82)?
P( p^ > 0.82) = _____ (Round four decimals places)
5) What is the probability of obtaining x = 111 of fewer individuals with the characteristic ? that is what is P(p^ < 0.74)?
P(p^ < 0.74) = _______ (Round to four decimal places as needed)
1) The correct phrase that best describes the shape of the sampling distribution of p^ is (b) Approximately normal because n < 0.05N and np (1-p) > 10.
2) The mean of the sampling distribution of p^ is equal to the population proportion, which is p. Therefore, p^ = 0.8.
3) The standard deviation of the sampling distribution of p^ is given by the formula sqrt(p * (1-p) / n). Plugging in the values, we have sqrt(0.8 * (1-0.8) / 150) = 0.0259.
4) To find the probability of obtaining x = 123 or more individuals with the characteristic, we need to calculate the z-score and find the corresponding area under the standard normal distribution curve. The formula for the z-score is (x - mean) / standard deviation. Plugging in the values, we have z = (0.82 - 0.8) / 0.0259 = 0.7735. Using a standard normal distribution table or calculator, we find that the area to the right of z = 0.7735 is approximately 0.2182.
5) To find the probability of obtaining x = 111 or fewer individuals with the characteristic, we again need to calculate the z-score and find the corresponding area under the standard normal distribution curve. The formula for the z-score is (x - mean) / standard deviation. Plugging in the values, we have z = (0.74 - 0.8) / 0.0259 = -2.3182. Using a standard normal distribution table or calculator, we find that the area to the left of z = -2.3182 is approximately 0.0103.
(1) The correct answer is b) Approximately normal because n ≤ 0.05N and np(1-p) > 10. This is because b) satisfies both conditions required for the sampling distribution of p^ to be approximately normal. The conditions are: (a) the sample size is less than or equal to 5% of the population size (n ≤ 0.05N), and (b) both np and n(1-p) are greater than 10.
To determine if the conditions are satisfied, we check if n ≤ 0.05N and np(1-p) > 10, which in this case are 150 ≤ 0.05 * 20000 and 150 * 0.8 * (1-0.8) > 10. Both conditions are satisfied, so the sampling distribution of p^ is approximately normal.
(2) To determine the mean of the sampling distribution of p^, we use the formula μp^ = p, which states that the mean of the sampling distribution of p^ is equal to the true population proportion p. Therefore, in this case, the mean of the sampling distribution of p^ is p^ = 0.8.
(3) To determine the standard deviation of the sampling distribution of p^, we use the formula σp^ = √((p * (1-p)) / n), which states that the standard deviation of the sampling distribution of p^ is equal to the square root of the product of p and (1-p), divided by the sample size n. Therefore, in this case, the standard deviation of the sampling distribution of p^ is σp^ = √((0.8 * (1-0.8)) / 150) ≈ 0.0278 (rounded to six decimal places).
(4) To calculate the probability of obtaining x=123 or more individuals with the characteristic (P(p^ > 0.82)), we use the Z-score formula. The Z-score formula is given by Z = (p^ - p) / σp^, where p^ is the observed sample proportion, p is the true population proportion, and σp^ is the standard deviation of the sampling distribution of p^. In this case, we have p^ = 0.82, p = 0.8, and σp^ ≈ 0.0278. Plugging these values into the formula, we have Z = (0.82 - 0.8) / 0.0278 ≈ 0.7194.
Next, we use a Z-table to find the probability associated with the Z-score of 0.7194, which is P(Z > 0.7194) ≈ 0.2358. Therefore, the probability of obtaining x=123 or more individuals with the specified characteristic is P(p^ > 0.82) ≈ 0.2358 (rounded to four decimal places).
(5) Similarly, to calculate the probability of obtaining x = 111 or fewer individuals with the characteristic (P(p^ < 0.74)), we use the Z-score formula again. Plugging in the values p^ = 0.74, p = 0.8, and σp^ ≈ 0.0278 into the formula, we have Z = (0.74 - 0.8) / 0.0278 ≈ -2.1583.
Using the Z-table, we find the probability associated with the Z-score of -2.1583, which is P(Z < -2.1583) ≈ 0.0152. Therefore, the probability of obtaining x=111 or fewer individuals with the specified characteristic is P(p^ < 0.74) ≈ 0.0152 (rounded to four decimal places).
(1)
The sample distribution taken from a binomial distribution is approximately normal because sample/population size < 0.05 ("small" sample). Also a binomial distribution approximates a normal distribution when np>K and np(1-p)>K, where K ranges between 5 & 10, depending on preference.
(2)
The expected value of p̂, or E[p̂] is the mean of the population. The mean of a binomial population is np.
(4)
P(X≥123)=ΣP(Xi),i=123...n
where P(Xi)=C(n,i)pi(1-p)(n-i)
and C(n,i)=n!/(i!(n-i)!) is combination of i objects from n.
It's a relative long process to do the summation. You can use binomial tables, or you can approximate the value using the normal approximation to binomial distribution, as hinted in part (1).
(5) see (4)