Question

Construct a 95% confidence interval of the population proportion using the given information x=180, n = 300 the lower bound is the upper bound isRound to three decimal places as needed

Answer 1

To construct a 95% confidence interval for a population proportion using x=180 and n=300, we can use the formula (p-hat)±Z√[(p-hat)(1-p-hat)/n], where p-hat is the sample proportion, Z is the Z-value for the desired confidence level, and n is the sample size. By substituting the given values into the formula, we find the confidence interval to be (0.810, 0.874).

Step-by-step explanation:

To construct a 95% confidence interval for a population proportion, we use the formula:

(p-hat) ± Z × √[(p-hat)(1-p-hat)/n]

where p-hat is the sample proportion, Z is the Z-value for the desired confidence level, and n is the sample size. Given x=180 and n=300, we can calculate p-hat as 180/300 = 0.6. The Z-value for a 95% confidence level is approximately 1.96. Substituting these values into the formula, we get:

(0.6) ± 1.96 × √[(0.6)(1-0.6)/300]

Simplifying the expression inside the square root yields:

(0.6) ± 1.96 × √0.0008

Calculating the square root and performing the multiplication, we get:

(0.6) ± 1.96 × 0.0283

Rounding to three decimal places, the confidence interval is (0.810, 0.874).

Answer 2

Given;

`x=180,n=300`

Then, we can find the point estimation as;

`\begin{gathered} \hat{p}=(x)/(n) \\ \hat{p}=(180)/(360)=0.60 \end{gathered}`
`Z_{(\alpha)/(2)}=Z_(0.05)=1.96`

Thus, the margin of error E is;

`\begin{gathered} E=Z_{(\alpha)/(2)}\sqrt[]{\frac{\hat{p}(1-\hat{p})}{n}} \\ E=1.96\sqrt[]{(0.60(0.40))/(300)} \\ E=1.96\sqrt[]{0.0008} \\ E=0.055 \end{gathered}`

A 95% confidence interval for population proportion p is;

`\hat{p}\pm E=0.60\pm0.055`

The lower bound is;

`0.60-0.055=0.545`

The upper bound is;

`0.60+0.055=0.655`