Chapter 6, Problem 10TP

Solution

To Find: The number of people surveyed.

The number of people surveyed is $38$

Given information: The interval is $45.99\%-78.01\%$ for the customers who prefer brand X cereal over brand Y.

Calculation:

1. Calculate the sample proportion.

  $\frac{0.4599+0.7801}{2}=0.62$

2. Assume that result obtained at $95\%$ confidence interval then, half of the interval is:

  $\frac{0.7801-0.4599}{2}=0.1601$ v

3. Now, use the student’s distribution as, the population information is not available to calculate the half of the interval:

  $t^{*}\sqrt{\frac{\widehat{p}\left(1-\widehat{p}\right)}{n}}=0.1601$

4. Use the maximum value of $t^{*}$ at $95\%$ confidence level of $1.96$ .

  $\begin{array}{l}\sqrt{\frac{(0.62)(0.38)}{n}}={\left(\frac{0.1601}{1.96}\right)}^2\\ n=\left(0.62\right)\left(0.38\right){\left(\frac{1.96}{0.1601}\right)}^2\\ n=35.3\end{array}$

5. When $n=35.3$ and the degree of freedom comes out to be $34$ and $t^{*}=2.03$ .
6. Recalculate to obtain:

  $\begin{array}{l}$ n=9.1916{(t^{*})}^2$n=9.1916{\left(2.03\right)}^2\\ n=37.9\end{array}$