BA 2 - Sampling and estimation

If the average height of all women is 63.5 inches and the standard deviation is 2.5 inches, approximately what percentage of women are between 58.5 and 68.5 inches tall?

95% is correct 58.5 and 68.5 inches are two standard deviations from the mean, that is 63.5±2(2.5). According to the rules of thumb, approximately 95% of women's heights fall within two standard deviations of the mean.

If the mean weight of all students in a class is 165 pounds with a variance of 234.09 square pounds, what is the z-value associated with a student whose weight is 140 pounds?

-1.63 is correct z=x−μ/σ=140−165/15.3≈−1.63z . The standard deviation, 15.3, is the square root of the variance, 234.09.

For a standard normal distribution (µ=0, σ=1), the area under the curve less than 1.25 is 0.894. What is the approximate percentage of the area under the curve less than -1.25?

0.106 is correct 1-0.894=0.106 is the area under the curve for all values greater than 1.25. Since the normal distribution is symmetric, 0.106 is also the area under the curve for all values less than -1.25.

If a particular standardized test has a mean score of 500 and standard deviation of 100, what percentage of test-takers score between 500 and 600?

100 is one standard deviation above the mean (600-500 =100= 1*100 = 1*stdev). We know that approximately 68% of the distribution is within 1 standard deviation of the mean. Therefore 34% must fall beyond 1 standard deviation above the mean.

According to the Central Limit Theorem, the means of random samples from which of the following distributions will be normally distributed, assuming the samples are sufficiently large?

All of the above is correct According to the Central Limit Theorem, if we take large enough samples, the distribution of sample means will be normally distributed regardless of the shape of the underlying population.

What happens to the sample mean and standard deviation as you increase the sample size?

The sample mean and standard deviation generally become closer to the population mean and standard deviation is correct As we increase the sample size, the sample includes more members of the population, so it is less likely to include only unusual values. Therefore, as the sample grows, the sample mean and standard deviation approach the population mean and standard deviation.

How large must our sample size be for the 95% confidence interval to be within 1 kg/m2 of the true average BMI? Since we don't know σσ, the standard deviation of the population, let's use the standard deviation of our previous sample (s=7.10) as an estimate.

n≥194 is correct If we use the equation n≥(zs/M)2to calculate n, we find n≥(1.967.101)2≥193.66n. However, because the sample size must be a whole number, we have to round up so n≥194.

What is the standard deviation of the distribution of sample means?

σ/√n, the population standard deviation divided by the square root of the sample size, is the standard deviation of the distribution of sample means. Large samples will create a "tighter" distribution of sample means than smaller samples.

Assuming the same level of confidence, how does the width of the confidence interval for small samples compare with that for large samples?

It is wider is correct Smaller samples have greater uncertainty, which means wider confidence intervals.

Recall that the z-value associated with a value measures the number of standard deviations the value is from the mean. Given that the average height of all women is 63.5 inches and the standard deviation is 2.5 inches, what z-value corresponds to 61 inches?

-1 is correct z=x−μ/σ, so z=61−63.5/2.5=−2.5/2.5=−1

If the average IQ is 100 and the standard deviation is 15, approximately what percentage of people have IQs above 130?

2.5% 130 is two standard deviations above the mean (130-100=30=2*15=2*stdev). We know that approximately 95% of the distribution is within 2 standard deviations of the mean. Therefore 5% must fall beyond 2 standard deviations, 2.5% at the top and 2.5% at the bottom.

The mean of women's heights is 63.5 inches and the standard deviation is 2.5 inches. Select the two heights that define the lower and upper bounds of a range that encompasses about 99.7% of all women's heights.

56 inches is correct A range that is three standard deviations from the mean encompasses about 99.7% of a normal distribution. In this case, the lower bound of the range equals the mean minus three standard deviations =63.5-3(2.5)=56 inches. 71 inches is correct A range that is three standard deviations from the mean encompasses about 99.7% of a normal distribution. In this case, the upper bound of the range equals the mean plus three standard deviations =63.5+3(2.5)=71 inches.

If the mean of a normally distributed population is -10 with a standard deviation of 2, what is the likelihood of obtaining a value less than or equal to -7?

=NORM.DIST(-7,-10,2,TRUE)

A researcher wants to select a random sample of consumers for a study. Generate a random ID number between 0 and 1,000 for each consumer in the spreadsheet.

=RAND()*1000

What is the probability of obtaining a value less than or equal to two standard deviations below the mean?

Approximately 2% is correct You can use the interactive to solve this problem. If you position the slider so that it highlights the range from the far left side to "z=−2z=−2", you can see that the area under the curve over than range is approximately 2%. This is the cumulative probability associated with z=−2z=−2. Therefore, the probability of obtaining a value less than or equal to two standard deviations below the mean is approximately 2%.

What probability falls within one standard deviation of the mean?

Approximately 68% is correct The phrase "within one standard deviation of the mean" means "between one standard deviation below the mean and one standard deviation above the mean." This answer can be found using the rules of thumb for the normal distribution or by using the previous interactive. 68% of the probability lies within one standard deviation of the mean.

You report a confidence interval to your boss but she says that she wants a narrower range. SELECT ALL of the ways you can reduce the width of the confidence interval.

Increase the sample size is correct Increasing the sample size provides a more accurate representation of the population and therefore, reduces the width of the confidence interval. Note that another option is also correct. Decrease the confidence level is correct Decreasing the confidence level reduces the width of the confidence interval. Note that another option is also correct

Which of the following would increase the width of the confidence interval? Select all that applly

Increasing the confidence level is correct Increasing the confidence level means that we must be more confident that the actual population mean lies within our range. The confidence level must be wider to increase the likelihood that it captures the true population mean. Note that confidence level determines the z-value, which in turn drives the width of the interval. Note that another option is also correct. Decreasing the sample size is correct Decreasing the sample size will result in a less accurate prediction, and, therefore, a wider confidence interval. Note that n is in the denominator, so as n decreases, s/√n increases, that is, the width of the confidence interval increases. Note that another option is also correct.

Which of the following is the MOST LIKELY result of using a survey with biased questions?

The data in your sample will differ in a systematic way from data based on unbiased random selections from the population. is correct In general, surveys with biased questions may lead to biased data, which differ systematically from what would be seen in an unbiased sample. For example, biased survey questions would lead to systematic differences between the answers given on your surveys and the answers that would be that would be given on a more neutral survey.

A journalist wants to determine the average annual salary of CEOs in the S&P 1,500. He does not have time to survey all 1,500 CEOs but wants to be 95% confident that his estimate is within $50,000 of the true mean. The journalist takes a preliminary sample and estimates that the standard deviation is approximately $449,300. What is the minimum number of CEOs that the journalist must survey to be within $50,000 of the true average annual salary? Remember that the z-value associated with a 95% confidence interval is 1.96. Please enter your answer as an integer; that is, as a whole number with no decimal point.

The formula for calculating the minimum required sample size is n≥(zs/M)2, where M=50.000 is the desired margin of error for the confidence interval, s=$449,300 is the sample standard deviation, and z=1.96. Using these data we find that (1.96*449,300/50,000)2=310.200 Since n must be an integer (let's not even think of what 0.20 CEOs would look like!) and n must be greater than or equal to 310.20, we must round up to 311. Since 310.20 is closer to 310 than to 311, we would normally round 310.20 down to 310. However, in this case we must round up to find the smallest integer that satisfies the equation. Therefore, the minimum required sample size is 311.

What is the center value of the distribution of the sample means?

The population mean (μμ) is correct According to the Central Limit Theorem, if we take enough large samples, the mean of the set of sample means equals the population mean.

Suppose that you have a sample with a mean of 50. You construct a 95% confidence interval and find that the lower and upper bounds are 42 and 58. What does this 95% confidence interval around the sample mean indicate? Select all that apply.

We are 95% confident that the population mean lies between 42 and 58. is correct The 95% confidence interval is a range around the sample mean. We can say that we are 95% confident that the true population mean is within this range, based on the methods we used to calculate the range. If we were to construct similar intervals for 100 samples drawn from this population, on average 95 of the intervals will contain the true population mean.

How large must our sample be for the 68% confidence interval to be within 1 kg/m2 of the true average BMI? Since we don't know σ, the standard deviation of the population, let's use the standard deviation of our previous sample (s=7.10) as an estimate.

n≥51is correct If we use the equation n≥(zs/M)2to calculate n, we find n≥(17.101)2≥50.41 However, because the sample size must be a whole number, we have to round up, so n≥51 We round up even though 0.41 is less than 0.50 because if 50.41 is the smallest value that satisfies the inequality, then 50 will not satisfy the equation.

Recall that the z-value associated with a value measures the number of standard deviations the value is from the mean. If a particular standardized test has an average score of 500 and a standard deviation of 100, what z-value corresponds to a score of 350?

z=(x-µ)/σ. Here x= 350, µ=500, the population mean, and σ=100, the population standard deviation. Thus z = (350-500)/100 = (-150)/100 = -1.5