Question

3. (8 pta) A researcher wishing to estimate the proportion of X-ray machines that malfunction and produce excess radiation. A random sample of 40 machine is taken and 12 of the machines malfunction Computea 98% confidence interval for the proportion of X-ray machine that malfunction. 4. (7 pts) A federal report indicated that 27% of the children ages 2 to 5 years had a good diet, an increase over previous year. How large a sample is needed to estimate the true proportion of children with good diets within 2% with 95% level of confidence?

Answer 1

Given that Out of 40 machines, 12 machines malfunctions. Sample size n = 40 sample proportion p = 20 = 0.3 Now we have to construct the 98% confidence interval for true population proportion (p). Step (1): 100 1- 98 = 1 -0.98 = 0.02 0,02 = 0.01 Step (2): 1 -0.01 = 0.99 Now we have to find the value of Za/2 such that P(Z < Za/2) = 0.99 From standard normal table Za/2 = 2.326 (Or) Use excel formula: ENORM.S.INV(0.99)

Step (3): First find the margin of error. P(1-P) margin of error = MOE = Za/2* Vn 0.3+(1-0.3) 2= 0.169 V40 :: margin of error = MOE = 0.169 Step (4): Lower bound= p - MOE = 0.3 -0.169 = 0.131 Upper bound = P + MOE = 0.3 +0.169 = 0.469 :: (0.131, 0.469) 4) Given that Sample proportion = p = 27% = 0.27 sample size = n Now we have to calculate the sample size 'n' for 95% confidence interval with margin of error MOE = 2% = 0.02.

Step (1): = 1-0.95 = 0.05 0.05 = 0.025 Step (2): 1 -0.025 = 0.975 Now we have to find the value of z such that P(Z < Za/2) = 0.975 From standard normal table Za/2 = 1.96 (Or) Use excel formula: =NORM.S.INV(0.975)

Step (3): MOE = Zarz - (02- = 34.59) = (CE) =n=() = p= (1-P) =n= (1-95)* * 0.27 + (1 – 0.27) 1 1 n = 1892.95 n 1893 in= 1893

thank you.