Question

A consumer advocate group selects a random sample of 205 snack foods and finds that 174 of them contain more fat grams than what is indicated in the snack's nutritional information. Create a 90% confidence interval for the population proportion for snack foods that contain more fat grams that what is indicated in the snack's nutritional information. Enter the lower and upper bounds for the interval in the following boxes, respectively. You may answer using decimals rounded to four places or a percentage rounded to two. Make sure to use a percent sign if you answer using a percentage.

Answer 1

Solution :

Given that,

n = 205

x = 174

Point estimate = sample proportion = = x / n = 174/205=0.849

1 -   = 1- 0.849 =0.151

At 90% confidence level the z is

= 1 - 90% = 1 - 0.90 = 0.1

/ 2 = 0.1 / 2 = 0.05

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

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

= 1.645 *((0.849*0.151) /205 )

E= 0.0411

A 90% confidence interval for population proportion p is ,

- E < p < + E

0.849- 0.0411< p <0.849+ 0.0411

0.8079< p < 0.8901

The 90% confidence interval for the population proportion p is : lower bound =0.8079,upper bound = 0.8901