sta 406/403 assignment solution spring 2018

Sta406

Solution

Q1.

Since there are two tails of the normal distribution, the 95% confidence level would imply the 97.5^th percentile of the normal distribution at the upper tail. Therefore, z_α∕2 is given by qnorm(.975).

> zstar = qnorm(.975) 
> p = 0.5 
> E = 0.05 
> zstar^2 ∗ p ∗ (1−p) / E^2 
[1] 384.15

Answer

With a planned proportion estimate of 50% at 95% confidence level, it needs a sample size of 385 to achieve a 5% margin of error for the survey of female student proportion.

Q2.

step 1: Subtract 1 from your sample size. 10 – 1 = 9. This gives you degrees of freedom, which you’ll need in step 3.

Step 2: Subtract the confidence level from 1, then divide by two. 
(1 – .95) / 2 = .025

Step 3: Look up your answers to step 1 and 2 in the t-distribution table. For 9 degrees of freedom (df) and α = 0.025, my result is 2.262.

Step 4: Divide your sample standard deviation by the square root of your sample size. 
25 / √(10) = 7.90569415

Step 5: Multiply step 3 by step 4.
2.262 × 7.90569415 = 17.8826802

Step 6: For the lower end of the range, subtract step 5 from the sample mean.
240 – 17.8826802 = 222.117

Step 7: For the upper end of the range, add step 5 to the sample mean. 
240 + 17.8826802 = 257.883