Question

Using the data in the file named chapter 12 data set 2, test the research hypothesis at the .05 level of significance that boys raise their hand in class more often than girls. do this practice problem by hand, using a calculator. what is your conclusion regarding the research hypothesis? remember to first decide whether this is a one- or two-tailed test. salkind, neil j.; frey, bruce b.. statistics for people who (think they) hate statistics (p. 252). sage publications. kindle edition.

Answer 1

It is observed that males raise their hands more frequently than females for answering the question.

What is the difference of means?

The mean difference is investigated to determine that whether values (directed toward that measure) of a two groups differ considerably or not, taking into account the equality in variance between given groups.

The Z test is employed based on the understanding about population data (here, standard deviation), while the t-test is used if sample values are utilized to estimate them.

Now, according to the question;

Then let population mean be the amount of men raising their hands $\mu_{M}$  as well as the population mean be the number of females raising their hands $\mu_{F}$.

Then the hypothesis would be;

$\begin{aligned}&H_{0}: \mu_{M} \leq \mu_{F} \\&H_{1}: \mu_{M} > \mu_{F}\end{aligned}$

So, the test shows the result of right trailed.

Calculate the sample mean for the men;

$\begin{aligned}\bar{x}_{M} &=\frac{\sum_{i=1}^{n} X_{i}}{n} \\&=\frac{9+8+4+9+3+8+10+8+9+8+10+7+6+12}{14} \\&=7.93\end{aligned}$

Then, the standard deviation for males will become,

$\begin{aligned}s_{M} &=\sqrt{\frac{\sum_{i=1}^{n}\left(X_{i}-\bar{x}_{M}\right)^{2}}{n-1}} \\&=\sqrt{\frac{(9-7.93)^{2}+(8-7.93)^{2}+(4-7.93)^{2}+\ldots+(7-7.93)^{2}+(6-7.93)^{2}+(12-7.93)^{2}}{14-1}} \\&=2.3685\end{aligned}$

Now, calculate the mean for females;

$\begin{aligned}\bar{x}_{F} &=\frac{\sum_{i=1}^{n} X_{i}}{n} \\&=\frac{3+5+1+2+6+4+3+6+7+9+7+3+7+6+8+8}{16} \\&=5.31\end{aligned}$

Then, the standard deviation for female will become;

$\begin{aligned}s_{F} &=\sqrt{\frac{\sum_{i=1}^{n}\left(X_{i}-\bar{x}_{F}\right)^{2}}{n-1}} \\&=\sqrt{\frac{(3-7.93)^{2}+(5-7.93)^{2}+(1-7.93)^{2}+\ldots+(6-7.93)^{2}+(8-7.93)^{2}+(8-7.93)^{2}}{16-1}} \\&=2.3866\end{aligned}$

Assuming variance equality, a pooled variance is:

$\begin{aligned}s_{p}^{2} &=\frac{\left(n_{M}-1\right) s_{M}^{2}+\left(n_{F}-1\right) s_{F}^{2}}{n_{M}+n_{F}-2} \\&=\frac{(14-1) 2.3685^{2}+(16-1) 2.3866^{2}}{14+16-2} \\&=5.6559\end{aligned}$

The, the test statistics would be-

$\begin{aligned}t &=\frac{\bar{x}_{M}-\bar{x}_{F}}{\sqrt{s_{p}^{2}\left(\frac{1}{n_{M}}+\frac{1}{n_{F}}\right)}} \\&=\frac{7.93-5.31}{\sqrt{5.6559\left(\frac{1}{14}+\frac{1}{16}\right)}} \\&=3.0103\end{aligned}$

The degree of freedom will become-

$\begin{aligned}d f &=n_{M}+n_{F}-2 \\&=14+16-2 \\&=28\end{aligned}$

The critical value is 1.701 at 28 degrees of freedom & 0.05 level of significance.

As, $t > t_{\text {critical }}$, the null hypothesis will be rejected.

As a result, males are more likely than females to raise their hands in response to the inquiry.

To know more about the null hypothesis, here

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The correct question-

Using the data set below, test the research hypothesis at the .05 level of significance that males raise their hands in class more often than females. Do this practice problem by hand using a calculator. What is your conclusion regarding the research hypothesis? Remember to first decide whether this is a one- or two-tailed test. (The table is attached)