Let X1, ...., Xn be a random sample from a normal population given by the normal density f(x) = 1 σ √2 2π e −(x−µ) 2 2σ2 , −∞ < x < ∞ 1. Construct for σ known the 90 percent confidence interval for the unknown mean µ by using the standard normal distributed random variable Z = √2 n(X − µ) σ with X = 1 n Pn i=1 Xi . (Hint:you may use that for a standard normal distribution P(Z ≤ 1.65) ≈ 0.95)
Let X1, ...., Xn be a random sample from a normal population given by the normal density f(x) = 1 σ √2 2π e −(x−µ) 2 2σ2 , −∞ < x < ∞
1. Construct for σ known the 90 percent confidence interval for the unknown mean µ by using the standard normal distributed random variable Z = √2 n(X − µ) σ with X = 1 n Pn i=1 Xi . (Hint:you may use that for a standard
2. Consider the random variable Z = √2 n(X − µ) S with S 2 = 1 n − 1 Xn i=1 (Xi − X) 2 the so-called sample variance and determine the cdf of this random variable. Construct now for σ unknown the 90 percent confidence interval for the unknown mean µ by using this random variable Z (Hint:you may use all the knowledge you know about normal distributions. In particular, you may use that the random variable (n−1)S 2 σ2 has a chisquare distribution with n − 1 degrees of freedom and for the random variable Z there exist some number tn−1,0.05 satisfying P(Z ≤ tn−1,0.05) = 0.95
3. If we consider two independent samples of size n1 and size n2 both from a normal population with unknown parameters µ1, σ1 and µ2, σ2 having the same unknown variance σ1 = σ2 = σ determine the distribution of the statistic Z = (X − Y ) − (µ1 − µ2) Sp 2 q 1 n1 + 1 n2 with S 2 p = (n1 − 1)S 2 1 + (n2 − 1)2S 2 2 n1 + n2 − 2 and S 2 i the sample variance of population i, i = 1, 2. (hint:You may use all the knowledge you know about chi-square distributions and normal populations without proof)
4. Construct a 90 percent confidence interval for the difference µ1 −µ2 by using the statistic defined in part 3.
5. Toyota manufacturers a variety of cars over a wide
6. In part 5 we applied the result of the previous three parts for a normal population. Is that realistic in this particular example? Explain!
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