Of 84 adults selected randomly from one town, 62 have health insurance. Find a 90% confidence interval for the true proportion of all adults in the town who have health insurance.
Please help mr solve.
To find the confidence interval for the true proportion of all adults in the town who have health insurance, you can use the formula for confidence intervals for proportions. Here are the steps to follow:
1. Define the variables:
- n: Number of adults randomly selected (84 in this case)
- x: Number of adults who have health insurance (62 in this case)
- p̂: Sample proportion (x/n)
- z: Z-score corresponding to the desired confidence level (90% in this case)
- E: Margin of error
2. Calculate the sample proportion:
p̂ = x / n
p̂ = 62 / 84
3. Find the z-score corresponding to the desired confidence level:
For a 90% confidence level, the z-score is 1.645. You can look up this value in a standard normal distribution table or use a calculator.
4. Calculate the margin of error:
E = z * √((p̂ * (1 - p̂)) / n)
E = 1.645 * √((0.738 * (1 - 0.738)) / 84)
5. Calculate the lower and upper limits of the confidence interval:
Lower Limit = p̂ - E
Upper Limit = p̂ + E
So, using the values we calculated above:
Lower Limit = p̂ - E
Upper Limit = p̂ + E
Plug in the values:
Lower Limit = 0.738 - E
Upper Limit = 0.738 + E
Now you can calculate the margin of error using the equation E = 1.645 * √((0.738 * (1 - 0.738)) / 84). The margin of error is approximately 0.076.
Finally, calculate the confidence interval:
Lower Limit = 0.738 - 0.076
Upper Limit = 0.738 + 0.076
Therefore, the 90% confidence interval for the true proportion of all adults in the town who have health insurance is approximately 0.662 to 0.814. This means we are 90% confident that the true proportion lies within this range.