Chapter 24, Problem 15E

To determine

To explain do these results suggest that the new activities are better and test an appropriate hypothesis and state your conclusion.

Answer to Problem 15E

There is a sufficient evidence to support the claim of an improvement.

Explanation of Solution

The scores of the school children are given in the question. So, by calculating the mean and standard deviation of the data using excel function we get,

The formulas used in excel to find the mean and standard deviations are as follows:

1. AVERAGE ( $\text{number1, }\left[\text{number2}\right]$, ...): Returns the average (arithmetic mean) of the arguments.
2. STDEV ( $\text{number1, }\left[\text{number2}\right]$,...): Estimates standard deviation based on a sample.

For the New activities:

New activities
Average	=AVERAGE(C1:C18)
S.D.	=STDEV(C1:C18)

And the result will be as:

New activities
Average	51.7222
S.D.	11.7062

Now, for the Control group:

Control
Average	=AVERAGE(D1:D22)
S.D.	=STDEV(D1:D22)

And the result will be:

Control
Average	41.8182
S.D.	16.5979

Thus, we have,

  $\begin{array}{l}{\overline{x}}_1=51.7222\\ {\overline{x}}_2=41.8182\\ n_1=18\\ n_2=22\\ s_1=11.7062\\ s_2=16.5979\end{array}$

Now, let us define the hypotheses as:

  $\begin{array}{l}{H}_0:{\mu }_1={\mu }_2\\ H_1:{\mu }_1>{\mu }_2\end{array}$

The value of test statistic will be:

  $\begin{array}{l}t=\frac{{\overline{x}}_1-{\overline{x}}_2}{\sqrt{\frac{s_1^2}{n_1}+\frac{s_2^2}{n_2}}}\\ =\frac{51.7222-41.8182}{\sqrt{\frac{{11.7062}^2}{18}+\frac{{16.5979}^2}{22}}}\\ =-8.94\end{array}$

The degrees of freedom will be:

  $df=37$

Then the P-value will be:

  $P<0.005$

Thus, if the P-value is less than or equal to the significance level, then the null hypothesis is rejected.

  $P<0.05\Rightarrow \text{Reject }{H}_0$

Thus, we conclude that there is a sufficient evidence to support the claim of an improvement.