Chapter 8: Q1E (page 452)

The purpose of this exercise is to compare the variability of ${\bar{x}}_{1} and {\bar{x}}_{2}$ with the variability of $({\bar{x}}_{1} - {\bar{x}}_{2})$ .
a. Suppose the first sample is selected from a population with mean $μ_{1} = 150$ and variance $σ_{1}^{2} = 900$ . Within what range should the sample mean vary about $95 %$ of the time in repeated samples of measurements from this distribution? That is, construct an interval extending standard deviations of $x_{1}$ on each side of $μ_{1}$ .
b. Suppose the second sample is selected independently of the first from a second population with mean $μ_{2} = 150$ and variance $σ_{2}^{2} = 1600$ . Within what range should the sample mean vary about $95 %$ the time in repeated samples of 100 measurements from this distribution? That is, construct an interval extending standard deviations $x_{2}$ on each side $μ_{2}$ .
c. Now consider the difference between the two sample means $({\bar{x}}_{1} - {\bar{x}}_{2})$ . What are the mean and standard deviation of the sampling distribution $({\bar{x}}_{1} - {\bar{x}}_{2})$ ?
d. Within what range should the difference in sample means vary about the $95 %$ time in repeated independent samples of $100$ measurements each from the two populations?
e. What, in general, can be said about the variability of the difference between independent sample means relative to the variability of the individual sample means?

Short Answer

Expert verified

The standard deviation is a statistic that calculates the square root of the variance as well as quantifies the dispersal of a collection compared to its average.

Step by step solution

Central Limit Theorem.

According to theCentral Limit Theorem,the sampling distribution of the sample means approaches a normal distribution, irrespective of the shape of population distribution if the sample size is over 30.

(a) Find the interval extending standard deviations x1 on each side μ1 .

It is given that $μ_{1} = 150, σ_{1}^{2} = 900, n = 100$ and confidence level is 95 % .

So, the confidence interval is $μ \pm 2 \sqrt{\frac{σ^{2}}{n}}$

$= 150 \pm 2 \sqrt{\frac{900}{100}} = 150 \pm 2 \sqrt{9} = 150 \pm (2 \times 3) = 150 \pm 6$

Therefore, the interval is from 144 to 156.

(b) Find the interval extending standard deviations x2 on each side μ2 .

It is given that $μ_{2} = 150, σ_{2}^{2} = 1600, n = 100$ and confidence level is 95% .

So, the confidence interval is $μ \pm 2 \sqrt{\frac{σ^{2}}{n}}$

$= 150 \pm 2 \sqrt{\frac{1600}{100}} = 150 \pm 2 \sqrt{16} = 150 \pm (2 \times 4) = 150 \pm 8$

Therefore, the interval is from 142 to 158 .

(c) Find the mean and standard deviation of the sampling distribution (x¯1- x¯2).

It is given that $μ_{1} = 150, μ_{2} = 150, σ_{1}^{2} = 900, σ_{2}^{2} = 1600 a n d n = 100$

$\begin{array}{rcl} x_{1} - x_{2} & = & μ_{1} - μ_{2} \\ = & 150 - 150 \\ = & 0 \end{array}$

$\begin{array}{rcl} σ_{x_{1} - x_{2}} & = & \sqrt{\frac{σ_{1}^{2}}{n_{1}} + \frac{σ_{2}^{2}}{n_{2}}} \\ = & \sqrt{\frac{900}{100} + \frac{1600}{100}} \\ = & \sqrt{9 + 16} \\ = & \sqrt{25} \\ = & 5 \end{array}$

Therefore, the mean and standard deviation of the sampling distribution $(x_{1} - x_{2})$ is 0 and 5 respectively.

(d) State the range within which the difference in sample means varies.

The difference in sample means will be normally distributed according to the Central Limit Theorem as $n > 30$ .

(e) State the conclusion.

In general, we can say that the difference in sample means for independent sampling has variability equal to the sum of the individual variability of the means. It can be expressed as:

$E (x_{1} - x_{2}) = μ_{d} = μ_{1} - μ_{2}$