- (10 points)Given a sample size of n = 484. Let the variance of the population be σ2 = 10.24. Let the mean of the sample be xbar = 20. Construct a 90% confidence interval for µ, the mean of the population, using this data and the central limit theorem.
Use Summary 5b, Table 1, Column 1
- What is the standard deviation (σ) of the population? = 3.2
- What is the standard deviation of the mean xbar of sample size n, i.e. what is σxbar , in terms of σ and n using the central limit theorem?
- Is this a one-sided or two-sided problem?
- What value of z should be used in computing k, the margin of error, where
z = k/σxbar ?
- What is k?
- Write the 90% confidence interval for µ based on xbar and k,
(xbar – k) < µ < (xbar + k)
- Using the Z-score applet “Area from a value”. Let the Mean = 20, and SD = σxbar. Choose “Between (xbar -k) and (xbar + k)” using xbar = 20 and your computed value of k. Hit “Recalculate”. Does the probability approximately equal 0.90? (yes or no). Include a screen shot of your answer.
- (10 points) Candidate A claims he is getting a 60% positive approval rating. A recent poll has candidate A polling a 55% approval rating based on a sample of 600 voters (meaning 330 voters said they were approving of candidate A). Construct a 95% proportion confidence interval based on this data and determine whether the data supports the 60% claim.
Start by using the binomial distribution ( + )n = 1, and = n.
Use Summary 5b, Table 1, Column 3.
- What is n? = 600
- What is ? =
- What is ? =
- What is ? =
- What is p? (Note: p is based on our belief about the population. is based on the sample.) =
- Let σp = sqrt(). What is σp ?
- We want to make a confidence interval by using the formula
z = k/σp
Should we use a 1 tail or 2 tail z-score?
- What value of z corresponds to the desired 95% confidence interval?
- What is k?
- Construct the confidence interval
( – k) < ptrue< ( + k)
- Is your confidence intervalconsistent with the belief that 60% approval rating claim? (yes or no, and explain your answer using your computed confidence interval).
| Type of sample | Population parameters | Sample statistics |
| General | µσ N | s n |
| Binomial (proportion) | p σ N | σ or s n |
Note: A “General” sample means that the values in the population can be any number between -inf and +inf. A “Binomial” sample means that the values in the population can take on only 2 values like: p/not p, heads/tails, black/white, male/female, on/off, 1/0, filled/empty, etc.
Table 1: Confidence Intervals
| σ is known – General sample | σ is unknown – General sample | σ is approximately known – Binomial sample |
| σ n | s n | σ = |
| z = k/[σ/] | t = k/[s/] | z = |
| k = z*σ/
where z = 1.645, 1.96, 2.576 (90%,95%,99%) | k = t*s/ find t with t-distribution applet using: 2 tails, df = (n – 1), p = .10, .05, .01 | k = z* where z = 1.645, 1.96, 2 .576 |
| ( – k) <µ< ( + k) | ( – k) <µ< ( + k) | ( – k) < p < ( + k)
|
Table 2: Statistical Comparison of Mean Values
| General sample vs theoretical value | General sample vs General sample | Binomial sample vs theoretical value | Binomial sample vs Binomial sample |
| Ho: µ = constant Ha: µ> constant | Ho: µ1 = µ2 Ha: µ1>µ2 | Ho: p = constant Ha: p > constant | Ho: p1 = p2 Ha: p1> p2 |
| s n | s1n1 s2 n2
s = | s = | s1 = s2 = s = |
| t = | t = | t = | t = |
| df = n – 1 right tail | df = (n1 – 1) + (n2 – 1) right tail | df = n -1 right tail | df = (n1 – 1) + (n2 – 1) right tail |
Using the t-distribution applet, find the probability p. (Don’t confuse this p with the proportion constant in Table 2, Column 3!)
Compare p with α.
If p >α, accept Ho.
If p <α, reject Ho, accept Ha instead.