Estimation Checkpoint
We say that a point estimator is unbiased if (choose one):
its sampling distribution is centered exactly at the parameter it estimates.
Due to a limited budget, the researcher obtained opinions from a random sample of only 2,222 U.S. adults. With this sample size, the researcher can be 95% confident that the obtained sample proportion will differ from the true proportion (p) by no more than (answers are rounded):
2.1% Remember that for a sample size of n, the sample proportion p ˆ will differ from the "true" population proportion by no more than 1/√n. In this case, this is about 2.1%. Confidence intervals are used to find a region in which we are 100 * ( 1 - a )% confident the true value of the parameter is in the interval. For large sample confidence intervals about the population proportion you have: pHat ± z * sqrt(phat * (1- phat) / n) where phat is the sample proportion z is the zscore for having a% of the data in the tails, i.e., P( |Z| > z)= a n is the sample size To find the sample size needed for a confidence interval of a given size we need only to concern ourselves with the error term and the width of the interval. We know that the interval is centered at phat so we need to find the value of n such that z * sqrt(phat * (1-phat) / n) = width. The z-score for a 0.95 confidence interval is the value of z such that 0.025 is in each tail of the distribution. z= 1.96 If we don't know anything about phat and are still asked to find the sample size we let phat = 0.5. This maximizes the value of the error term and if n is sufficient for phat = 0.5, the n will be sufficient for all other values of phat. z * sqrt(phat * (1-phat) / n) = width. 1.96 * sqrt( 0.50 * (1 - 0.50) / 2222) = 0.02078998
The next four questions refer to the following information: A study was conducted in order to estimate μ, the mean number of weekly hours that U.S. adults use computers at home. Suppose a random sample of 81 U.S. adults gives a mean weekly computer usage time of 8.5 hours and that from prior studies, the population standard deviation is assumed to be σ = 3.6 hours.
Based on this information, what would be the point estimate for μ? 8.5 The point estimate for the population mean μ is the sample mean, x ¯ . In this case, to estimate the mean number of weekly hours of home-computer use among the population of U.S. adults, we used the sample mean obtained from the sample, therefore x ¯ = 8.5.
Which of the following will provide a more informative (i.e., narrower) confidence interval than the one in problem 3?
Both using a sample of size 400 (instead of 81) and using a 90% level of confidence (instead of 95%) are correct. In general, we can obtain a more informative (narrower) confidence interval in one of two ways: compromising on the level of confidence (in other words, choosing a lower level of confidence), or increasing the sample size (that might be impossible in practice, though).
These next two questions refer to the following information: A researcher would like to estimate p, the proportion of U.S. adults who support recognizing civil unions between gay or lesbian couples. If the researcher would like to be 95% sure that the obtained sample proportion would be within 1.5% of p (the proportion in the entire population of U.S. adults), what sample size should be used?
In order to estimate the population proportion, p, with a 95% confidence interval with a margin of error of m, we need a sample size of (at least) 1/m2. In this case, the desired m is 1.5% = .015. Thus, the required n is 4,445 (remember, always round up for sample size). 1/n(squared) so 1/.015 squared = 4444.4444 = B 4445
How large a sample of U.S. adults is needed in order to estimate μ with a 95% confidence interval of length 1.2 hours?
We would like our confidence interval to be a 95% confidence interval (implying that z* = 2) and the confidence interval length should be 1.2, therefore the margin of error (m) = 1.2 / 2 = .6. The sample size we need in order to obtain this is: 144.
We are 95% confident that the mean number of weekly hours that U.S. adults use computers at home is:
between 7.7 and 9.3. The 95% confidence interval for the mean, μ, is x ¯ ± 2 ⋅ σ n = 8.5 ± 2 ⋅ 3.6 81 = 8.5 ± . 8 = ( 7.7 , 9.3 ) . .
