95% z-confidence interval for the proportion of all children enrolled in kindergarten who attended preschool is between a lower limit of 0.528 and an upper limit of 0.772.

Step-by-step explanation:

Confidence interval = p + or - zsqrt[p(1-p) ÷ n]

p is sample proportion = 39/60 = 0.65

n is the number of children sampled = 60

Confidence level (C) = 95% = 0.95

Significance level = 1 - C = 1 - 0.95 = 0.05

Divide significance level by 2 to obtain critical value (z)

0.05/2 = 0.025 = 2.5%

z at 2.5% significance level = 1.96

zsqrt[p(1-p) ÷ n] = 1.96sqrt[0.65(1-0.65) ÷ 60] = 1.96sqrt[0.2275 ÷ 60] = 1.96sqrt(3.792×10^-3) = 1.96×0.062 = 0.122

Lower limit = p - 0.122 = 0.65 - 0.122 = 0.528

Upper limit = p + 0.122 = 0.65 + 0.122 = 0.772

95% confidence interval is between 0.528 and 0.772

5 0

3 years ago

Claire is considering investing in a new business. In the first year, there is a probability of 0.2 that the new business will l

tangare [24]

The expected value of business profit is the sum of probability times profit:

... 0.2×(-10,000) + 0.4×0 + 0.3×5,000 + 0.1×8,000

... = -2000 + 1500 + 800

... = 300 . . . dollars

The expected value of the business return is slightly positve, so Claire should invest in the business.

8 0

3 years ago