The proportion of adults living in a small town who are college graduates is estimated to be p = 0.6. To test this hypothesis, a random sample of 15 adults is selected. If the number of college graduates in the sample is anywhere from 6 to 12, we shall not reject the null hypothesis that p = 0.6; otherwise, we shall conclude that p �/= 0.6.
Evaluate β for the alternatives p = 0.5 and p = 0.7.
Beta is The probability of Type 2 error.
I tried to use the binomial distribution with n=15 and p=.5 and .7 to get P(6<=x<=12|p=.5) and use 12 and 6 for x and substract the results but I can't get the correct answers which are 0.8454 for p=0.5, and 0.8695 for p=0.7
Is there another way of doing it? Are my x's right for the binomial distribution right?
β = P(Type II Error) =P(6≤X ≤13|p=0.5)
(0.5)^x(0.5)^15-x ≈ 0.8454
β = P(Type II Error) =P(6≤X ≤13|p=0.7
0.7^x (0.3)^15-x ≈
you're right
To evaluate β for the alternatives p = 0.5 and p = 0.7, you can use the binomial distribution to calculate the probability of observing 6 to 12 successes (college graduates) in a random sample of 15 adults.
Let's first calculate β for p = 0.5.
Using the binomial distribution with n = 15 (sample size) and p = 0.5 (null hypothesis), you need to find the probability of observing 6 to 12 successes (college graduates).
β for p = 0.5 can be calculated as follows:
β = P(6 ≤ x ≤ 12 | p = 0.5)
= P(x = 6) + P(x = 7) + P(x = 8) + P(x = 9) + P(x = 10) + P(x = 11) + P(x = 12)
To calculate these probabilities, you can use a binomial probability calculator or formula:
P(x = k) = nCk * p^k * (1-p)^(n-k)
Where nCk represents the number of combinations of n items taken k at a time.
Now, for p = 0.5:
β = P(x = 6) + P(x = 7) + P(x = 8) + P(x = 9) + P(x = 10) + P(x = 11) + P(x = 12)
= (15C6 * 0.5^6 * (1-0.5)^9) + (15C7 * 0.5^7 * (1-0.5)^8) + (15C8 * 0.5^8 * (1-0.5)^7) + (15C9 * 0.5^9 * (1-0.5)^6) + (15C10 * 0.5^10 * (1-0.5)^5) + (15C11 * 0.5^11 * (1-0.5)^4) + (15C12 * 0.5^12 * (1-0.5)^3)
Evaluating the above expression will give you β = 0.8454 (rounded to four decimal places), which matches the correct answer you provided.
Now, let's calculate β for p = 0.7.
Using the same approach, you can calculate β for p = 0.7:
β = P(6 ≤ x ≤ 12 | p = 0.7)
= P(x = 6) + P(x = 7) + P(x = 8) + P(x = 9) + P(x = 10) + P(x = 11) + P(x = 12)
= (15C6 * 0.7^6 * (1-0.7)^9) + (15C7 * 0.7^7 * (1-0.7)^8) + (15C8 * 0.7^8 * (1-0.7)^7) + (15C9 * 0.7^9 * (1-0.7)^6) + (15C10 * 0.7^10 * (1-0.7)^5) + (15C11 * 0.7^11 * (1-0.7)^4) + (15C12 * 0.7^12 * (1-0.7)^3)
Evaluating the above expression will give you β = 0.8695, which matches the correct answer you provided.
So, the correct values for β are 0.8454 for p = 0.5 and 0.8695 for p = 0.7.
To evaluate β for the alternatives p = 0.5 and p = 0.7, we need to calculate the probability of making a Type 2 error, which is the probability of rejecting the null hypothesis when it is actually true.
In this case, we are given that the null hypothesis is p = 0.6, and the alternative hypotheses are p = 0.5 and p = 0.7.
To calculate β, we need to find the probability of observing a sample mean proportion falling outside the range of 6 to 12, assuming that the true proportion is either 0.5 or 0.7.
To calculate β using the binomial distribution, we can use the following steps:
Step 1: Calculate the probability of observing a sample mean proportion falling within the specified range of 6 to 12 for each alternative value of p.
For p = 0.5:
P(6 ≤ x ≤ 12 | p = 0.5) = P(x = 6) + P(x = 7) + ... + P(x = 12)
= Σ(i=6 to 12) P(x = i) (using binomial distribution formula)
For p = 0.7:
P(6 ≤ x ≤ 12 | p = 0.7) = P(x = 6) + P(x = 7) + ... + P(x = 12)
= Σ(i=6 to 12) P(x = i) (using binomial distribution formula)
Step 2: Subtract the probabilities obtained in Step 1 from 1 to find β.
β = 1 - P(6 ≤ x ≤ 12 | p = 0.5) (for p = 0.5)
β = 1 - P(6 ≤ x ≤ 12 | p = 0.7) (for p = 0.7)
Now, let's calculate β for p = 0.5 and p = 0.7 using the binomial distribution:
For p = 0.5:
P(6 ≤ x ≤ 12 | p = 0.5) = P(x = 6) + P(x = 7) + ... + P(x = 12)
= Σ(i=6 to 12) [ C(15, i) * (0.5)^i * (0.5)^(15-i) ]
Calculating these probabilities using a binomial calculator or statistical software, we get:
P(6 ≤ x ≤ 12 | p = 0.5) ≈ 0.8454
Therefore, β for p = 0.5 is approximately 0.8454.
For p = 0.7:
P(6 ≤ x ≤ 12 | p = 0.7) = P(x = 6) + P(x = 7) + ... + P(x = 12)
= Σ(i=6 to 12) [ C(15, i) * (0.7)^i * (0.3)^(15-i) ]
Calculating these probabilities using a binomial calculator or statistical software, we get:
P(6 ≤ x ≤ 12 | p = 0.7) ≈ 0.8695
Therefore, β for p = 0.7 is approximately 0.8695.
By using the correct values for the binomial distribution and calculating the probabilities accordingly, we obtain the expected values for β as stated.