Question

It is generally believed that nearsightedness affects about 15% of children. A school district gives vision tests to 111 incoming kindergarten children. In our sample of 111 students, we find 13% of the students were nearsighted. Construct a 90% confidence interval for the number of nearsighted kindergarteners we would expect to see based on our sample. Does this support or refute the estimate of 15%? Assume conditions are met (so don't check them)! Explain in details

Answer 1

solution Giren P=013 na IV 2012 for 90% confidence = 1645 90.4 (I = P + Zal2 , pll-P) ñ = 0:13 1645 **.01370.87 =0:13 645* / 0:1131 0.13 +1645+0031920 = 0.13 + 0.052509 10.07749, 0:182509) Since 0:15 Lil in the intuval so, it does not refute The estimate of 15t.