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What is the sampling distribution of a statistic? Why is it important?

Short Answer

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A statistic's sampling distribution is the distribution of values obtained from all potential samples of the same size from the same population.

The sampling distribution of a statistic is used to determine the likelihood that the statistic's value is similar to other possible sample values. The value of statistics assists us in determining the likelihood of inaccuracy in calculating the population parameter.

Step by step solution

01

Explanation

The unknown parameters of the distribution determine the sampling distribution of a statistic. Although any observation is feasible within the distribution's range, we can rule out values with a low probability (or probability density) as being unlikely. As a result, it is acceptable to assume that the parameter values are close to the density's maximum.

Following the observations, the known values can be substituted into the density function. The only thing left is a function of the parameters. The probability function is the name for this. The inference is based on this function in the three main schools of thought: frequentists, Bayesians, and likelihoodlums.

Frequentists can make probabilistic judgments using the sampling distribution. If you choose how to generate a 95% confidence interval before collecting data, for example, it will have a 95% probability of having the correct value. We can decide whether the data has true value after we acquire it. Because we don't know which, people think of the parameter as having a 95% chance of being in the interval. Despite being incorrect, this is most likely harmless.

Only the likelihood is important to Bayesians and likelihoodlums. Intervals are used by Bayesians to indicate degrees of belief. However, they must assume a probability distribution for the parameters before collecting the data in order to create a 95 percent credible interval, which is an interval in which our degree of conviction that the parameter is in the interval is a 95 percent.

Such arbitrary assumptions irritate likelihoodlums and frequentists. Likelihoodlums don't employ a probability interpretation and instead rely solely on the likelihood function.

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Most popular questions from this chapter

7.16 NBA Champs. This exercise requires that you have done Exercises 7.11-7.15.
a. Draw a graph similar to that shown in Fig. 7.3 on page 294for sample sizes of 1,2,3,4, and 5.
b. What does your graph in part (a) illustrate about the impact of increasing sample size on sampling error?
c. Construct a table similar to Table 7.4 on page 294 for some values of your choice.

Population data: 2,5,8

Part (a): Find the mean, ฮผof the variable.

Part (b): For each of the possible sample sizes, construct a table similar to Table 7.2on the page 293and draw a dotplot for the sampling for the sampling distribution of the sample mean similar to Fig 7.1on page 293.

Part (c): Construct a graph similar to Fig 7.3and interpret your results.

Part (d): For each of the possible sample sizes, find the probability that the sample mean will equal the population mean.

Part (e): For each of the possible sample sizes, find the probability that the sampling error made in estimating the population mean by the sample mean will be0.5or less, that is, that the absolute value of the difference between the sample mean and the population mean is at most 0.5.

New York City 10-kmRun. As reported by Rumะปะต's World magazine, the times of the finishers in the New York City 10-km run are normally distributed with a mean of 61 minutes and a standard deviation of 9minutes. Do the following for the variable "finishing time61min of finishers in the New York City 10-kmrun.

a. Find the sampling distribution of the sample mean for samples of size 4

b. Repeat part (a) for samples of size 9

C. Construct graphs similar to those shown in Fig.7.4on-page 304

d. Obtain the percentage of all samples of four finishers that have mean finishing times within 5minutes of the population mean finishing time of 61 minutes. Interpret your answer in terms of sampling error.

e. Repeat part (d) for samples of size 9

Nurses and Hospital Stays. In the article "A Multifactorial Intervention Program Reduces the Duration of Delirium. Length of Hospitalization, and Mortality in Delirious Patients (Journal of the American Geriatrics Society, Vol. 53. No. 4. pp. 622-628), M. Lundstrom et al. investigated whether education programs for nurses improve the outcomes for their older patients. The standard deviation of the lengths of hospital stay on the intervention ward is 8.3days.

a. For the variable "length of hospital stay," determine the sampling distribution of the sample mean for samples of 80patients on the intervention ward.

b. The distribution of the length of hospital stay is right-skewed. Does this invalidate your result in part (a)? Explain your answer.

c. Obtain the probability that the sampling error made in estimating the population means length of stay on the intervention ward by the mean length of stay of a sample of 80patients will be at most 2days.

7.48 Menopause in Mexico. In the article "Age at Menopause in Puebla. Mexico" (Human Biology, Vol. 75, No, 2, Pp. 205-206), authors L. Sievert and S. Hautaniemi compared the age of menopause for different populations. Menopause, the last menstrual period, is a universal phenomenon among females. According to the article, the mean age of menopause, surgical or natural, in Puebla, Mexico is 44.8years with a standard deviation of 5.87years. Let x~denote the mean age of menopause for a sample of females in Puebla, Mexico.
a. For samples of size 40, find the mean and standard deviation of xยฏ. Interpret your results in words.
b. Repeat part (a) with n=120.

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