Question

Which expression should you simplify to find the 90% confidence interval,

given a sample of 36 people with a sample proportion of 0.45?

O

0.45(1-0.45)

A. 0.45 36. V - 1.645

O

B. 0.45 + 1.645 • V-

0.45(1-0.45)

36

36

O C. 0.45 + 1.645 • V 0.45(1–0.45)

D. 0.45 1.645,80.45

36

Answer 1

0.45 +/- 1.645 x (sq rt 0.45 [1-0.45] ) / 36

Explanation:

ur welcome

Answer 2

`0.45 - 1.96\sqrt{(0.45(1-0.45))/(36)}=0.2870.45 + 1.96\sqrt{(0.45(1-0.45))/(36)}=0.613`

we can conclude at 95% of confidence that the true proportion of interest for this case is between 0.287 and 0.613

Explanation:

The estimated proportion of interest is
`\hat p=0.45`

We need to find a critical value for the confidence interval using the normla standard distributon. For this case we have 95% of confidence, then the significance level would be given by
`\alpha=1-0.95=0.05` and
`\alpha/2 =0.025`.

And the critical value is:
`z_(\alpha/2)=-1.96, z_(1-\alpha/2)=1.96`

The confidence interval for the true population proportion is interest is given by this formula:

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

Replacing the values provided we got:

`0.45 - 1.96\sqrt{(0.45(1-0.45))/(36)}=0.287\\\\0.45 + 1.96\sqrt{(0.45(1-0.45))/(36)}=0.613`

And we can conclude at 95% of confidence that the true proportion of interest for this case is between 0.287 and 0.613