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Chapter 10: Q. 10.120 (page 441)

$H_{2} * μ_{1} \neq μ_{2}$

Short Answer

Expert verified

Since the value of test statistics is fall in the accepted region. Thus, the null hypothesis is not rejected.

Step by step solution

01

Given information

The hypothesis is

$H_{0} : μ_{1} = μ_{2}, H_{a} : μ_{1} = μ_{2}$

and the level of significance is 0

02

Explanation

Let's compute the sample mean as follow

$\bar{d} = \frac{\sum d}{n}$

$= \frac{7}{8}$

$= 0.875$

Calculate the value of standard deviation is,

$s_{d} = \sqrt{\frac{\sum d_{i}^{2} - \frac{{(\sum d_{i})}^{2}}{n}}{n - 1}}$

$= \sqrt{\frac{47 - \frac{(7)^{2}}{8}}{8 - 1}} = 2.4165$

The test statistics are,

$t = \frac{\bar{d}}{\frac{s_{d}}{\sqrt{n}}}$

Substitute the given values in the above equation.

$z = \frac{0.875}{\frac{2.4165}{\sqrt{5}}}$

$= 1.024$

The nvalue of degree of freedom is,

$d o f = n - 1 = 8 - 1 = 7$

The critical value for a level of significance of $0.1$ is $1.895$ .

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Most popular questions from this chapter

$H_{2} : μ_{1} > μ_{2}$

The Federal Bureau of Prisons publishes data in Prison Statistics on the times served by prisoners released from federal institutions for the first time. Independent random samples of released prisoners in the fraud and firearms offense categories yielded the following information on time served, in months.

At the $5 %$ significance level, do the data provide sufficient evidence to conclude that the meantime served for fraud is less than that for firearms offenses? (Note: ${\bar{x}}_{1} = 10.12, s_{1} = 4.90, {\bar{x}}_{2} = 18.78$ , and $s_{2} = 4.64$ .)

The primary concern is deciding whether the mean of Population 1 differs from the mean of Population 2 .

A variable of two population has a mean of $7.9$ and standard deviation of $5.4$ for one of the population and a mean of $7.1$ and a standard deviation of $4.6$ for the other population.

a. For independent samples of sizes $3 a n d 6$ respectively find the mean and standard deviation of $x_{1} - x_{2}$

b. Must the variable under consideration be normally distributed on each of the two population for you to answer part (a) ? Explain your answer.

b. Can you conclude that the variable $x_{1} - x_{2}$ is normally distributed? Explain your answer.

Driving Distances. Refer to Exercise 10.48 and determine a 95% confidence interval for the difference between last year's mean VMTs by midwestern and southern households.

See all solutions

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