Question

DistributionsIdentify the distribution (normal, Studentt,chi-square) that should be used in each of the following situations. If none of the three distributions can be used, what other method could be used?

a. In constructing a confidence interval of$\mu$, you have 75 sample values and they appear to be from a population with a skewed distribution. The population standard deviation is not known.

b. In constructing a confidence interval estimate of $\mu$, you have 75 sample values and they appear to be from a population with a skewed distribution. The population standard deviation is known to be 18.2 cm.

c. In constructing a confidence interval estimate of$\sigma$, you have 75 sample values and they appear to be from a population with a skewed distribution.

d. In constructing a confidence interval estimate of $\sigma$, you have 75 sample values and they appear to be from a population with a normal distribution.

e. In constructing a confidence interval estimate ofp,you have 1200 survey respondents and5% of them answered “yes” to the first question.

Short answer

a. Student’s t-distribution

b. Normal distribution

c. None of the distribution

d. Chi-square distribution

e. Normal distribution

Step-by-step solution

Given information

The distributions that are provided are normal, Student’s t, and chi square.

Define the distributions

Student t distribution: This distribution is applied in the case of a small sample size and unavailability of population standard deviation.

Normal distribution: The distribution is applied in the case of large sample size and availability of population standard deviation.

Chi-square distribution: This distribution is applied in the case of a normally distributed population and in the case of categorical data.

Identify the distribution

a.

The following information is given:

The size of the sample is 75.

The population standard deviation is not known.

The population follows the skewed distribution.

Since the population standard deviation is not known, and the sample values (n > 30) are selected from a population with a skewed distribution, the distribution that can be applied in the provided scenario is Student’s t.

Assumptions for Student’s t-distribution are as follows:

1) The population standard deviation is not known.

2) The population follows the skewed distribution.

Therefore, Student’s t-distribution is appropriate to use.

b.

The following information is given:

The size of the sample is 75.

The population standard deviation is 18.2.

The population follows the skewed distribution.

Since the population standard deviation (18.2) is known, and the sample size is greater than 30 , the distribution that can be applied in the provided scenario is the normal distribution.

Assumptions for normal distribution are as follows:

1) The population standard deviation is known.

2) The size of the sample must be large.

Therefore, the normal distribution is appropriate to use.

c.

The following information is given:

The size of the sample is 75.

The population follows the skewed distribution.

In the provided scenario, none of the three distributions is appropriate to use. However, a confidence interval could be computed using bootstrap methods as the distribution of the population is skewed.

Therefore, none of the distributions is appropriate to use.

d.

The following information is given:

The size of the sample is 75.

The population follows the normal distribution.

The chi-square distribution is suitable for the provided scenario as the population is normally distributed, and the confidence interval estimate for the standard deviation is to be constructed.

Therefore, the Chi-square distribution is appropriate to use.

e.

The following information is given:

The size of the sample is 1200.

Out of 1200, 5% of them answered “yes” to the first question.

The suitable distribution for the provided scenario is normal distribution as the sample size is greater than 30, that is,$1200⩾30$.

The below conditions are also satisfied:

$$\begin{gathered} np=1200\times 0.05 \\ =60 \\ ⩾5 \end{gathered}$$

$$\begin{gathered} n\left(1-p\right)=1200\times 0.95 \\ =1140 \\ ⩾5 \end{gathered}$$

Therefore, the normal distribution is appropriate to use.