Chapter 10: Q. 5 (page 665)

A researcher wished to compare the average amount of time spent in extracurricular activities by high school students in a suburban school district with that in a school district of a large city. The researcher obtained an SRS of $60$ high school students in a large suburban school district and found the mean time spent in extracurricular activities per week to be $6$ hours with a standard deviation of $3$ hours. The researcher also obtained an independent SRS of $40$ high school students in a large city school district and found the mean time spent in extracurricular activities per week to be $5$ hours with a standard deviation of $2$ hours. Suppose that the researcher decides to carry out a significance test of $H_{0 :} μ_{unbartum} = μ_{c i n}$ versus a two-sided alternative. T10.5. The correct test statistic is
(a) $t = \frac{(6 - 5) - 0}{\sqrt{\frac{3}{60} + \frac{2}{40}}}$
(b) $t = \frac{(6 - 5) - 0}{\sqrt{\frac{3^{2}}{60} + \frac{2^{2}}{40}}}$
(c) $t = \frac{(6 - 5) - 0}{\frac{9}{60} + \frac{3}{40}}$
(d) $z = \frac{(6 - 5) - 0}{\sqrt{\frac{3}{60} + \frac{2}{40}}}$
(e) $z = \frac{(6 - 5) - 0}{\sqrt{\frac{3^{2}}{60} + \frac{2^{2}}{40}}}$

Short Answer

Expert verified

The test statistic is option (b) $t = \frac{(6 - 5) - 0}{\sqrt{\frac{3^{2}}{60} + \frac{2^{2}}{40}}} .$

Step by step solution

Given information

The first number of high school is $60$

Mean is $6$

Standard deviation is $3$

The second number of high schools is $40$

Mean is $5$

Standard deviation is $2$ .

Explanation

The given data is

$x_{1} = 6$

$s_{1} = 3$

$n_{1} = 60$

$x_{2} = 5$

$s_{2} = 2$

$n_{2} = 40$

We need to utilize the two-sample t-test because we want to compare the differences between two population means without knowing the standard deviations.

Find the test statistic

localid="1650388304454" $t = \frac{({\bar{x}}_{1} - {\bar{x}}_{2}) - (μ_{1} - μ_{2})}{\sqrt{\frac{s_{1}^{2}}{n_{1}} + \frac{s_{2}^{2}}{n_{2}}}} = \frac{(6 - 5) - 0}{\sqrt{\frac{3^{2}}{60} + \frac{2^{2}}{40}}}$