I have the following two multiple choice questions that are really throwing me for a loop:
1. Based on 500 people a researcher calculates the 95% confidence interval for the population proportion p: 0.123<p<0.181.
a. There is a 95% chance that the true value of p lies between 0.113 and 0.171.
b. If many diff. samples of 300 were selected and a confidence interval was constructed based on that sample, in the long run, 95% of the confidence intervals would contain the true value of p.
c. If 100 different samples of 300 were selected and a confidence interval was constructed based each sample, exactly 99 of these confidence intervals contain the true value of p.
---In all honesty, none of these answers feels right, any thoughts?
2. Suppose all values in a data set are converted to z-scores. Which of the following would be true?
a. The mean and the standard deviations of the z-scores will be the same as the original data.
b. The mean of the z-scores will be zero and the standard deviation will be one.
c. Both the mean and the standard deviation of the z-scores will be zero.
---My instinct on this question is a, just because it is the only one that makes some sense. What do you think?
Thank you,
Joanne
With a 95% confidence interval, one can be 95% confident that the true value of p is contained within that interval.
If you calculate the mean and standard deviation of the z-scores, the mean of the z-distribution will be zero and the standard deviation will be one.
I hope this will help.
1. Based on the information provided, the correct answer is b. If many different samples of 300 were selected and a confidence interval was constructed based on each sample, in the long run, 95% of the confidence intervals would contain the true value of p.
To understand why option b is the correct answer, let's break down the concepts involved. In a confidence interval, we are trying to estimate a population parameter (in this case, the population proportion p) based on a sample. The confidence level (in this case, 95%) represents the percentage of confidence intervals that will contain the true population parameter.
In this scenario, a researcher calculated a 95% confidence interval based on a sample of 500 people and found that 0.123 < p < 0.181. This means that 95% of the time, if we repeated the sampling process and calculated a confidence interval, the true value of p would fall within this range.
Option a is incorrect because the confidence interval (0.123 < p < 0.181) does not represent a chance or probability. It is a statement about the range within which we can reasonably estimate the true value of p.
Option c is incorrect because it states that exactly 99 out of 100 different samples of 300 would contain the true value of p. While it is true that 95% of the confidence intervals would contain the true value of p, there is no guarantee that exactly 99 out of 100 intervals would do so.
2. The correct answer is b. The mean of the z-scores will be zero, and the standard deviation will be one.
When we convert a data set to z-scores, we calculate the number of standard deviations away from the mean each individual data point is. The formula for calculating z-scores is:
z = (x - μ) / σ
Where z is the z-score, x is the individual data point, μ is the mean of the original data set, and σ is the standard deviation of the original data set.
Converting data to z-scores does not change the shape of the distribution, but it standardizes it. This means that the mean of the z-scores will always be zero, indicating that on average, the z-scores are "centered" around zero. The standard deviation of the z-scores will always be one, indicating that they are measured in terms of standard deviations from the mean.
Option a is incorrect because the mean and standard deviation of the z-scores will not be the same as the original data set. The z-scores are standardized, and their mean and standard deviation will be different.
Option c is incorrect because only the mean of the z-scores will be zero, not the standard deviation.
I hope this clarification helps! Let me know if you have any further questions.