Answer:
7 is y= 0.25p+5
8 is 60t=1350
9 is y=5t+40
10 is y=1 1/4b
Step-by-step explanation:
Answer:
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
