Chapter 18, Problem 38SE

A survey asked the following question: Do you favor or oppose providing tax-funded vouchers or tax deductions to parents who send their children to private schools? Of the 2010 individuals surveyed, 905 favored the proposal, 1045 opposed the proposal, and 60 offered no opinion. Do the data indicate a significant difference in the preferences for the financial support of parents who send their children to private schools? Use a .05 level of significance.

To determine

Check whether there is significant difference in the preferences for the financial support of parents who send their children to private schools or not.

Answer to Problem 38SE

There is significant difference in the preferences for the financial support of parents who send their children to private schools.

Explanation of Solution

Calculation:

There are total of 2010 individuals, of which 905 favored the proposal, 1,045 opposed the proposal and 60 offered no opinion. The level of significance is 0.05.

The proportion of who favour the proposal is denoted as p.

The hypotheses are given below:

Null hypothesis:

$H_0:p=0.50$.

That is, there is no difference in the preferences for the financial support of parents who send their children to private schools.

Alternative hypothesis:

$H_a\text{: }p\ne 0.50$.

That is, there is a difference in the preferences for the financial support of parents who send their children to private schools.

From the data, there are $905+1,045=1,950$ adults responded.

For a sign test, the mean is $\mu =0.5n$, standard deviation is, $\sigma =\sqrt{0.25n}$ and the sample size is, $n=1,950$.

The mean is,

$\begin{array}{c}\mu =0.5\left(1,950\right)\\ =975\end{array}$

Thus, the mean is 975.

The standard deviation is,

$\begin{array}{c}\sigma =\sqrt{0.25\left(1,950\right)}\\ =\sqrt{487.5}\\ =22.0794\end{array}$

Thus, the standard deviation is 22.0794.

The probability of 905 plus signs in the lower tail can be obtained by using the continuity correction factor and normal approximation. Hence, the p-value is obtained by using normal distribution with $\mu =975\text{and }\sigma =22.0794$.

$\begin{array}{c}P\left(\text{905 or few plus signs}\right)=P\left(x\le 905.5\right)\\ =P\left(z\le \frac{905.5-975}{22.0794}\right)\\ =P\left(z\le -3.15\right)\end{array}$

Procedure:

Step by step procedure to obtain the above probability using Table 1 of Appendix B is given below:

• Locate the value –3.1 in the column, named z.
• Move towards the right along the row of –3.1, till the column named 0.05 is reached.
• The cell at the intersection of the row –3.1 and the column 0.05 gives the cumulative probability corresponding to the standard normal variable value -3.15.

Thus, $P\left(z\le -3.15\right)=0.0010$

That is,

$P\left(\text{905 or few plus signs}\right)=0.0010$.

The two tailed p-value is, $2\left(0.0010\right)=0.002$.

Rejection rule:

If $p\text{-value}\le \alpha$, then reject the null hypothesis $\left(H_a\right)$.

Conclusion:

Here the level of significance $\alpha =0.05$.

Here, $p\text{-value}\left(=0.002\right)\le \alpha \left(=0.05\right)$.

That is, p-value is less than significance level.

Therefore, the null hypothesis is rejected.

Thus, there is evidence to support the claim that “there is significant difference in the preferences for the financial support of parents who send their children to private schools”.