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In: Statistics and Probability

Use for questions 7-10: A random sample of 384 low income Vermonters found that 223 said...

Use for questions 7-10: A random sample of 384 low income Vermonters found that 223 said that lack of reliable transportation was an important factor that prevented them from finding a higher paying job. 9. Calculate the upper limit of a 95% confidence interval for the proportion of low-income Vermonters that believe that lack of reliable transportation is an important factor that prevents them from finding a higher paying job. (Round to three decimal places.)

Solutions

Expert Solution

Solution :

Given that,

n = 384

x = 223

Point estimate = sample proportion = = x / n = 223/384=0.581

1 -   = 1- 0.581 =0.419

At 95% confidence level the z is ,

= 1 - 95% = 1 - 0.95 = 0.05

/ 2 = 0.05 / 2 = 0.025

Z/2 = Z0.025 = 1.96   ( Using z table )

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

= 1.96 (((0.581*0.419) /584 )

E = 0.040

A 95% confidence interval for population proportion p is ,

- E < p < + E

0.581-0.040 < p <0.581+ 0.040

0.541< p < 0.621

The 95% confidence interval for the population proportion p is : 0.541, 0.621


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