Question

In our Condor Cafe, the Oxnard College Culinary Program offers the most delicious and affortable breakfast and lunch choices among the 152 Community Colleges in California. This week, a random sample of 186 orders were recorded, and 57 orders were for the $5-meals. Please find the 99.9% confidence interval for the true proportion of the orders for the $5-meals.

A. Enter your answer as an open-interval (i.e., parentheses) using decimals (not percents) accurate to three decimal places.

Confidence interval =

B. xpress the same answer as a tri-linear inequality using decimals (not percents) accurate to three decimal places.

< p <

C. Express the same answer using the point estimate and margin of error. Give your answers as decimals, to three places.

p =    ±

Answer 1

Solution :

Given that,

A) n = 186

x = 57

Point estimate = sample proportion = = x / n = 57 / 186 = 0.306

1 - = 1 - 0.306 = 0.694

At 99.9% confidence level

= 1 - 99.9%

=1 - 0.999 =0.001

/2 = 0.0005

Z/2 = Z_0.0005 = 3.291

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

E = 3.291 (((0.306 * 0.694) / 186)

E = 0.111

A 95% confidence interval for population proportion p is

B) - E < p < + E

0.306 - 0.111 < p < 0.306 + 0.111

( 0.195 < p < 0.417 )

C)   ± E  

= 0.306 ± 0.111

= ( 0.195, 0.417 )