Question

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 $H_0:{\mu }_{suburban}={\mu }_{city}$versus a two-sided alternative. Which is the correct standardized test statistic ?

(a)$z=\frac{(6-5)-0}{\sqrt{{\displaystyle \frac{3}{60}}+{\displaystyle \frac{2}{40}}}}$

(b) $z=\frac{(6-5)-0}{\sqrt{{\displaystyle \frac{3^2}{60}}+{\displaystyle \frac{2^2}{40}}}}$

(c) role="math" localid="1654192807425" $t=\frac{(6-5)-0}{{\displaystyle \frac{3}{\sqrt{60}}}+{\displaystyle \frac{2}{\sqrt{40}}}}$

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

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

Short answer

The correct answer is:

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

Step-by-step solution

Given information

We are given,

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

$s_1=3$

$n_1=60$

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

$s_2=2$

$n_2=40$

Explanation

Since we want to test the difference between the two population means, while the standard deviations are unknown, we need to use the two-sample $t$test.

Determine the test statistic:

$t=\frac{({\overline{x}}_1-{\overline{x}}_2)-({\mu }_1-{\mu }_2)}{\sqrt{{\displaystyle \frac{{s_1}^2}{n_1}}+{\displaystyle \frac{{s_2}^2}{n_2}}}}=\frac{(6-5)-0}{\sqrt{{\displaystyle \frac{3^2}{60}}+{\displaystyle \frac{2^2}{40}}}}$