Question

Grace is sampling the family harvest of tomatoes. She randomly selects 359 tomatoes and inspects them for cracking. Of the 359 tomatoes sampled, 119 of them were cracked. Construct a 99% Confidence interval of the proportion of tomatoes suffering from cracks. Answer with the upper bound only. Round to 4 decimal places, as needed.

Answer 1

Solution :

Given that,

n = 359

x = 119

Point estimate = sample proportion = = x / n = 119/359=0.331

1 -   = 1- 0.331 =0.669

At 99% confidence level the z is ,

  = 1 - 99% = 1 - 0.99 = 0.01

/ 2 = 0.01 / 2 = 0.005

Z/2 = Z0.005 = 2.576 ( Using z table )

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

= 2.576* (((0.331*0.669) / 359)

= 0.0640

A 99% confidence interval for population proportion p is ,

- E < p < + E

0.331-0.0640 < p < 0.331+ 0.0640

0.2670< p < 0.3950

The 99% confidence interval for the population proportion p is : 0.2670, 0.3950