Question

A research company is interested in determining the true proportion of Americans that work remotely. In a random sample of 200 individuals, 4.5% of them work remotely. Find a 90% confidence interval for the true proportion of Americans that work remotely.

Answer 1

solution:

Given data:

n = 200

Point estimate = sample proportion = = 4.5%=0.045

1 -   = 1- 0.045 =0.955

At 90% confidence level

= 1 - 90%  

= 1 - 0.90 =0.10

/2 = 0.05

Z/2 = Z0.05 = 1.645 ( Using z table )

Margin of error = E Z/2 *(( * (1 - )) / n)

= 1.645 *((0.045*0.955) / 200)

= 0.024

A 90% confidence interval for true proportion p is ,

- E < p < + E

0.045-0.024 < p < 0.045+0.024

0.021< p < 0.069

The 90% confidence interval for the true proportion p is : (0.021 , 0.069)

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