Question

(1 point) A government official is in charge of allocating social programs throughout the city of Vancouver. He will decide where these social outreach programs should be located based on the percentage of residents living below the poverty line in each region of the city. He takes a simple random sample of 130 people living in Gastown and finds that 22 have an annual income that is below the poverty line.

Part i) The proportion of the 130 people who are living below the poverty line, 22/130, is a:

A. parameter.
B. statistic.
C. variable of interest.

Part ii) Use the sample data to compute a 95% confidence interval for the true proportion of Gastown residents living below the poverty line.

(Please carry answers to at least six decimal places in intermediate steps. Give your final answer to the nearest three decimal places).

95% confidence interval

Answer 1

solution

Part i)

#B.Statistic

Option B is corect

Solution :

n = 130

x = 122

= x / n = 122 / 130= 0.938462

1 - = 1 - 0. = 0.061538

At 95% confidence level the z is ,

= 1 - 95% = 1 - 0.95 = 0.05

/ 2 = 0.05 / 2 = 0.025

Z_/2 = Z_0.025 = 1.96

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

= 1.96 * (((0.938462* 0.061538) / 130)

= 0.012107

A 95% confidence interval for population proportion p is ,

- E < P < + E

0.938462- 0.012107 < p < 0.938462+ 0.01210

0.926354< p < 0.950569

The 95% confidence interval for the population proportion p is : (0.926354 , 0.950569)