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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$H_{0:}{渭}_{\text{unbartum}}={渭}_{cin}$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)$z=\frac{(6-5)-0}{\sqrt{\frac{3^2}{60}+\frac{2^2}{40}}}$

Step by step solution

01

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$2$.

02

Explanation

The given data is

${\overline{x}}_1=6$

$s_1=3$

$n_1=60$

${\overline{x}}_2=5$

$s_2=2$

$n_2=40$

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" $$\begin{gathered} t=\frac{\left({\stackrel{炉}{x}}_1-{\stackrel{炉}{x}}_2\right)-\left({渭}_1-{渭}_2\right)}{\sqrt{\frac{s_1^2}{n_1}+\frac{s_2^2}{n_2}}} \\ =\frac{(6-5)-0}{\sqrt{\frac{3^2}{60}+\frac{2^2}{40}}} \end{gathered}$$

$t=\frac{(6-5)-0}{\sqrt{\frac{3^2}{60}+\frac{2^2}{40}}}$.

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