A market research expert assumes that response rate in a poll is p=20 A large volume of n=400 questionnaires will be sent to randomly selected prospective customers. Variable (Y) denotes the total number of those who would respond. The sample proportion hatp=Y/n is going to be anaylzed
1. Determine parameters of approximate distribution for p
Answer:{90,8}
2. Evaluate the chance that response rate would exceed 17.5%
??
3. Find approximate upper and lower fifth percentiles for p.
??
I am not sure how to do 2 and 3
To answer questions 2 and 3, we need to use the normal approximation to the binomial distribution since the sample size is large (n = 400).
2. To evaluate the chance that the response rate would exceed 17.5%, we need to find the probability of getting a value for the sample proportion (hatp) greater than 0.175.
First, we calculate the mean of the distribution using the assumed response rate: µ = p * n = 20% * 400 = 80.
Next, we find the standard deviation (σ) of the distribution: σ = sqrt(p * (1 - p) / n) = sqrt(0.20 * 0.80 / 400) = 0.02.
Now, we can standardize the value of 0.175 using the z-score formula:
z = (0.175 - µ) / σ = (0.175 - 0.20) / 0.02 = -1.25.
Using a z-table or a statistical software, we find the area under the standard normal curve to the right of -1.25, which represents the probability of the response rate exceeding 17.5%.
The probability of the response rate exceeding 17.5% is equivalent to the area under the curve to the right of -1.25, which is approximately 0.8944 (or 89.44%).
3. To find the approximate upper and lower fifth percentiles for p, we need to find the values of p that correspond to the 95th percentile (or 0.95) and the 5th percentile (or 0.05).
First, we find the z-scores associated with the two percentiles. For the 95th percentile, we look for the z-score that corresponds to an area of 0.95 to the left of it. Similarly, for the 5th percentile, we look for the z-score that corresponds to an area of 0.05 to the left of it.
Using a z-table or a statistical software, we find that for the 95th percentile, the z-score is approximately 1.645, and for the 5th percentile, the z-score is approximately -1.645.
To find the upper fifth percentile, we use the formula: x = µ + (z * σ) = 0.20 + (1.645 * 0.02) = 0.2339 (or 23.39%).
To find the lower fifth percentile, we use the same formula, but with the negative value of the z-score: x = µ + (-1.645 * σ) = 0.20 + (-1.645 * 0.02) = 0.1661 (or 16.61%).
Therefore, the approximate upper fifth percentile for p is 23.39%, and the approximate lower fifth percentile for p is 16.61%.