Question

The First National Bank of Wilson has 600 checking account customers. Recent sample of 50 of these customers show 29 have a visa card with the bank. Construct the 90% confidence interval for the proportion of checking account customers who have a visa card with the bank. (Use a Z distribution table0

Answer 1

Answer:

$0.58 - 1.64\sqrt{\frac{0.58(1-0.58)}{50}}=0.466$

$0.58 + 1.64\sqrt{\frac{0.58(1-0.58)}{50}}=0.694$

The 90% confidence interval would be given by (0.466;0.694)

Step-by-step explanation:

The estimated proportion of interest would be:

$\hat p=\frac{29}{50}= 0.58$

In order to find the critical value we need to take in count that we are finding the interval for a proportion, so on this case we need to use the z distribution. Since our interval is at 90% of confidence, our significance level would be given by $\alpha=1-0.90=0.1$ and $\alpha/2 =0.05$. And the critical value would be given by:

$z_{\alpha/2}=-1.64, z_{1-\alpha/2}=1.64$

The confidence interval for the mean is given by the following formula:  

$\hat p \pm z_{\alpha/2}\sqrt{\frac{\hat p (1-\hat p)}{n}}$

If we replace the values obtained we got:

$0.58 - 1.64\sqrt{\frac{0.58(1-0.58)}{50}}=0.466$

$0.58 + 1.64\sqrt{\frac{0.58(1-0.58)}{50}}=0.694$

The 90% confidence interval would be given by (0.466;0.694)