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

`0.58 - 1.64\sqrt{(0.58(1-0.58))/(50)}=0.466`

`0.58 + 1.64\sqrt{(0.58(1-0.58))/(50)}=0.694`

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

Explanation:

The estimated proportion of interest would be:

`\hat p=(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{(\hat p (1-\hat p))/(n)}`

If we replace the values obtained we got:

`0.58 - 1.64\sqrt{(0.58(1-0.58))/(50)}=0.466`

`0.58 + 1.64\sqrt{(0.58(1-0.58))/(50)}=0.694`

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