The mean of the sampling distribution of x-bar equals the population mean mu.
Correct Answer
True
True. Sample means are centered at the population mean, so the mean of all possible values of x-bar is mu for any sample size.
Category: Sampling Distribution of Means
Question 2
The standard deviation of the sampling distribution of x-bar equals sigma divided by n.
Correct Answer
False
False. The standard deviation of x-bar, the standard error, is sigma divided by the square root of n, not by n itself.
Category: Standard Error
Question 3
The central limit theorem applies even when the population is not normal, provided the sample size is large enough.
Correct Answer
True
True. The power of the theorem is that sample means become approximately normal as n grows, whatever the shape of the population they come from.
Category: Sampling Distribution of Means
Question 4
As the sample size increases, the distribution of sample means becomes more spread out.
Correct Answer
False
False. The standard error sigma / sqrt(n) gets smaller as n increases, so the distribution of sample means becomes narrower, not more spread out.
Category: Standard Error
Question 5
The standard deviation of the sum of n values from a population is sqrt(n) times sigma.
Correct Answer
True
True. By the central limit theorem for sums, a total of n values has standard deviation sqrt(n) times sigma, which grows more slowly than n.
Category: CLT for Sums
Question 6
The mean of the sum of n values from a population equals mu divided by n.
Correct Answer
False
False. The mean of a sum of n values is n times mu. It is the sample mean, not the sum, whose mean equals mu.
Category: CLT for Sums
Question 7
If a population is normally distributed, its sample means are normally distributed for any sample size.
Correct Answer
True
True. Averages of values from a normal population are themselves exactly normal, so no large sample is needed in that case. The large-sample requirement matters for populations that are not normal.
Category: Sampling Distribution of Means
Question 8
The central limit theorem says that as the sample size grows, the population itself becomes more nearly normal.
Correct Answer
False
False. The population's shape never changes with sampling. The theorem says the distribution of the sample means becomes approximately normal as n grows.
Category: Sampling Distribution of Means
Question 9
For a population with mu = 40 and sigma = 6, the standard error of the mean for samples of n = 9 is 2.
Correct Answer
True
True. The standard error is sigma / sqrt(n) = 6 / sqrt(9) = 6 / 3 = 2. The population mean does not affect the standard error.
Category: Standard Error
Question 10
For a binomial experiment with n = 50 and p = 0.2, the mean of the approximating normal distribution is 25.
Correct Answer
False
False. The mean is np = 50 x 0.2 = 10. A mean of 25 would require p = 0.5.
Category: Normal Approximation to Binomial
Question 11
A binomial count can be approximated by a normal distribution when n is large and p is not too close to 0 or 1.
Correct Answer
True
True. Under those conditions the binomial distribution is nearly symmetric and bell-shaped, so a normal curve with mean np and standard deviation sqrt(npq) approximates it well.
Category: Normal Approximation to Binomial
Question 12
The standard error of the mean depends on the sample size but not on the population standard deviation.
Correct Answer
False
False. The standard error is sigma / sqrt(n), so it depends on both. A more variable population gives a larger standard error for the same sample size.
Category: Standard Error
Question 13
Sample means from samples of size n greater than 1 vary less than the individual values in that population.
Correct Answer
True
True. Averaging lets high and low values partly cancel, so the standard deviation of sample means, sigma / sqrt(n), is smaller than sigma whenever n is greater than 1. With n = 1, each sample mean is just one individual value, so the two spreads would be equal.
Category: Means Versus Individuals
Question 14
Averaging n values makes the variance of x-bar n times as large as the variance of a single value.
Correct Answer
False
False. The variance of x-bar is the population variance divided by n, sigma^2 / n. Taking its square root gives the standard error sigma / sqrt(n).
Category: Standard Error
Question 15
The probability distribution of all possible values of x-bar for a given sample size is called the sampling distribution of the sample mean.
Correct Answer
True
True. A statistic such as x-bar changes from sample to sample, and its distribution across all possible samples of size n is its sampling distribution.
Category: Sampling Distribution of Means
Question 16
For a population with mu = 10 and sigma = 5 and samples of n = 25, a sample mean of 12 has a z-score of 0.4.
Correct Answer
False
False. The standard error is 5 / sqrt(25) = 1, so z = (12 - 10) / 1 = 2. The value 0.4 is what a single value of 12 would score, using sigma instead of the standard error.
Category: Z-Scores and Percentiles
Question 17
When a normal curve approximates a binomial, P(X = 8) is found as the area between 7.5 and 8.5.
Correct Answer
True
True. The continuity correction treats the whole number 8 as the interval from 7.5 to 8.5, because a single point has zero area under a continuous curve.
Category: Normal Approximation to Binomial
Question 18
With the sample size held fixed, a population with a larger sigma produces a smaller standard error of the mean.
Correct Answer
False
False. The standard error sigma / sqrt(n) is proportional to sigma, so a larger population standard deviation produces a larger standard error.
Category: Standard Error
Question 19
When a fair coin is tossed 400 times, the normal curve that approximates the number of heads has a standard deviation of 10.
Correct Answer
True
True. The standard deviation is sqrt(npq) = sqrt(400 x 0.5 x 0.5) = sqrt(100) = 10.
Category: Normal Approximation to Binomial
Question 20
Samples of 64 drawn from a population with standard deviation 16 have a standard error of 0.25.
Correct Answer
False
False. The standard error is 16 / sqrt(64) = 16 / 8 = 2. The value 0.25 comes from dividing sigma by n instead of by the square root of n.
Category: Standard Error