In: Statistics and Probability
Recently, the number of children Americans have has dropped. The Gallup Poll organization wanted ... Recently, the number of children Americans have has dropped. The Gallup Poll organization wanted to know whether this was a change in attitude or possibly a result of the recession, so they asked Americans whether they have children, dont have children but wish to, or dont want children. Out of a random sample of 1200 American adults in 2013, 1140 said that they either have children or want/wish to.
(a) Suppose the margin of error in the confidence interval above were 0.02. What would that represent? The proportion of people in our sample who lied or misunderstood the question is 0.02. We actually called 1224 people; however, since 2% of them chose not to participate in our poll, we have an effective sample size of only 1200.
Our prediction is that the proportion of all Americans that have or want children is no more than 0.02 away from 0.95, the difference being due to randomness.
The difference between the actual proportion of Americans that have or want children and our sample proportion because of people lying or misunderstanding the question is 0.02.
The typical distance between the proportion of all Americans that have or want children and a sample proportion from a sample of size 1200 is 0.0
(b) If the margin of error were 0.02, the 95% confidence interval would be (0.93, 0.97). How would we interpret this interval in context?
I am 95% confident that the proportion of all Americans who said they have or want children is between 0.93 and 0.97.
In repeated sampling, 95% of the time, we would create an interval that captures the sample proportion of Americans who said they have or want children. This is one such interval.
There is a 95% chance that the proportion of Americans in the sample who said they have or want children is between 0.93 and 0.97.
There is a 95% chance that the proportion of all Americans who said they have or want children is between 0.93 and 0.97.
I am 95% confident that the proportion of Americans in the sample who said they have or want children is between 0.93 and 0.97.
(c) In 1990, 94% of all Americans either had children or wanted children. Has the proportion of all Americans who have or wish they had children changed since then? (Use the confidence interval (0.93, 0.97) for simplicity.) Why or why not?
Yes, because proportions are not the same from one sample to the next.
Yes, because 1140/1200 = 0.95 is not equal to 0.94. No, because 0.94 is in the confidence interval above.
Yes, because 0.94 is in the confidence interval above.
No, because 0.94 and 0.95 are not very far apart.
(a) Suppose the margin of error in the confidence interval above were 0.02. What would that represent?
The margin of error represents the difference between the sample proportion and lower limit of the confidence interval or the difference between upper limit of the confidence interval and the sample proportion.
Sample proportion = 1140/1200 = 0.95
Thus, the correct answer is,
Our prediction is that the proportion of all Americans that have or want children is no more than 0.02 away from 0.95, the difference being due to randomness.
(b) If the margin of error were 0.02, the 95% confidence interval would be (0.93, 0.97). How would we interpret this interval in context?
Interpretation of 95% confidence interval is - We're 95% confident that the interval (0.93, 0.97) captured the true proportion of all Americans that have or want children.
Thus, the correct answer is,
I am 95% confident that the proportion of all Americans who said they have or want children is between 0.93 and 0.97.
(c) In 1990, 94% of all Americans either had children or wanted children. Has the proportion of all Americans who have or wish they had children changed since then? (Use the confidence interval (0.93, 0.97) for simplicity.) Why or why not?
Since the value 94% lies in the confidence interval, there is no significant evidence that the proportion of all Americans who have or wish they had children changed since then.
Thus, the correct answer is,
No, because 0.94 is in the confidence interval above.