A large jar contains a mixture of white and black beans. In a randomly chosen handful of 120 beans 40 were black. Create a 98% confidence interval for the proportion of black beans in the jar. Express your answer to the nearest percent.

23%<x<43%

Step-by-step explanation:

The formula for calculating confidence interval is expressed;

Confidence Interval = p ± z√p(1-p)/n

n is the sample size

z is the z score at 98% confidence interval

n is the sample size = 120

z =  2.33

If in a randomly chosen handful of 120 beans 40 were black, then;

p = 40/120

p = 4/12

p = 0.33

CI = 0.33± 2.33√0.33(1-0.33)/120

CI = 0.33± [2.33√0.33(0.67)/120]

CI =  0.33± [2.33√0.2211/120]

CI = 0.33± [2.33(0.04292)]

CI = 0.33±0.100014

CI = (0.33-0.100014, 0.33+0.100014)

CI = (0.229, 0.430014)

CI = (22.9, 43.0014)

Hence 98% confidence interval for the proportion of black beans in the jar is 23%<x<43%.

i dont have an answer

step-by-step explanation: