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Chapter 10: Q. 10.121 (page 442)

$H_{a} * μ_{1} < μ_{2}$

Short Answer

Expert verified

Since the value of the test statistic is fall in the rejection region. Thus, the null hypothesis is 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 $0.1$ .

02

Explanation

Let's compute the sample mean as follow

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

$= \frac{- 21}{9}$

$= - 2.333$ .

Compute the value of standard deviation:

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

$= \sqrt{\frac{121 - \frac{(- 21)^{2}}{9}}{9 - 1}} = 3$

The test statistics are,

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

Substitute the given values in the above equation.

$t = \frac{- 2.333}{\frac{9}{\sqrt{9}}} = - 2.333$

The value of degree of freedom is,

$d o f = n - 1 = 9 - 1 = 8$

The critical value for the level of significance of $0.1$ is $- 1.397 .$

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Most popular questions from this chapter

A variable of two populations has a mean of $7.9$ and a standard deviation of $5.4$ for one of the populations and a mean of $7.1$ and a standard deviation of $4.6$ for the other population. Moreover. the variable is normally distributed in each of the two populations.

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

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

c. Determine the percentage of all pairs of independent samples of sizes $4$ and $16$ , respectively, from the two populations with the property that the difference $x_{1} - x_{2}$ between the simple means is between $- 3$ and $4$ .

Discuss the basic strategy for comparing the means of two populations based on independent simple random samples.

In each of Exercises 10.75-10.80, we have provided summary statistics for independent simple random samples from two populations. In each case, use the non pooled $t -$ fest and the non pooled t-interval procedure to conduct the required hypothesis test and obtain the specified confidence interval.

${\tilde{x}}_{1} = 20,$ $s_{1} = 2,$ $n_{1} = 30,$ $x_{2} = 18,$ $s_{2} = 5,$ $n_{2} = 40$ .

a. Right-tailed test, $α = 0.05$

b. $90 %$ confidence interval.

In each of exercise 10.13-10.18, we have presented a confidence interval for the difference, $μ_{1} - μ_{2}$ , between two population means. interpret each confidence interval

$99 %$ CI from $- 10 t o 5$

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

See all solutions

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