BS1501 tutorial 3_final

Applied Statistics and Mathematics in Economics & Business BS1501 Tutorial 3 Pairach Piboonrungroj (Champ) me@Pairach.com

1. A Yogurt manufacturer  x0 − µ (a) P ( x > x0 ) = P z >    σx   x0 = 570, µ = 500, σ x = 33  570 − 500 P ( x > 570) = P z >   33  = P ( Z > 2.12) = 0.017

1. A Yogurt manufacturer(b)  x0 − µ  P ( x > x0 ) = P z >    σx   σ 33 ~ ~ N ( µ = 500, σ ~ = 8.82)σx = = = 8.82 x x n 14  490 − 500 510 − 500  P (490 < x < 510) = P <Z <   8.82 8.82  = P(−1.13 < Z < 1.13) = 1 − 2 P ( z > 1.13) = 1 − 2(0.12924) = 0.740

Z = 1.13Z = -1.13

2. Women shift-workers(a) P(1 − P) = (0.60)(0.40) = 0.04 E ( p ) = 0.60 ˆ σp = ˆ n 150 ˆ  0.60 − 0.60 0.70 − 0.60  P (0.60 < P < 0.70) = P ( <Z<   0.04 0.04  P(0 < Z < 2.5) = 0.5 − 0.0062 = 0.4938(b) ˆ ≥ 0.50) = P Z > 0.50 − 0.60  P( P    0.04  P ( Z > −2.5) = 1 − P ( Z > 2.5) = 1 − 0.0062 = 0.9938

3. A Building Society s C.I. for µ = x ± Z α * 2 n s(a) 90% C.I. for µ = x ± 1.645 * n 8500 = 15200 ± 1.645 * 30 12667.16 < µ < 17752.84(b) s 8500 99 % C.I. for µ = x ± 2.57 * = 15200 ± 2.57 * n 30 11211.67 < µ < 19188.33

4. Voters from the electorate(a) 180 indicated their preference for a particular candidate (x) 400 sample voters (n) ˆ x P= n ˆ 180 P= = 0.45 400

4. Voters from the electorate(b) ˆ ˆ P(1 - P) ˆ C.I. for P = P ± Z α 2 n ˆ ˆ P(1 - P) ˆ95 % C.I. for P = P ± Z 0.05 2 n 0.45 * 0.55 = 0.45 ± 1.96 400 = 0.45 ± 0.049 0.401 < P < 0.499

4. Voters from the electorate(c) ˆ ˆ P(1 - P) ˆ C.I. for P = P ± Z α 2 n ˆ ˆ P(1 - P) ˆ99 % C.I. for P = P ± Z 0.01 2 n 0.45 * 0.55 = 0.45 ± 2.57 = 0.45 ± 0.064 400 0.386 < P < 0.514

5. Lifetime of light bulbs(a) n = 100, x = 1570, s = 120H 0 : µ = 1600 α = 0.05 ⇒ Reject H0 if z < -1.96 or if Z>+1.96 ⇒H 1 : µ ≠ 1600 α = 0.01 Reject H0 if z < -2.58 or if Z>+2.58 x−µ 1570 − 1600z= = = −2.5 < -1.96 BUT > -2.58 s 120 n 100 ∴Reject H0 at 5% level of significance i.e. There is sufficient evidence at 5% level of significance to reject the hypothesis that the mean lifetime of bulbs has changed, but not at the 1% level.

5. Lifetime of light bulbs (b)n = 100, x = 1570, s = 120 x − µ 1570 − 1600 z= = = −2.5 H 0 : µ = 1600 s 120 n 100 H1 : µ < 1600 α = 0.05 ⇒ Reject H0 if z < -1.64 (one tailed) Reject H0 if z < -2.33 (one α = 0.01 ⇒ tailed)∴Reject H0 5% and 1% levels of significance. The first alternative hypothesis would seem to be a fairly weak one compared to the second one, given that it would be of more interest to know whether the mean lifetime of the population of bulbs has deteriorated, rather than simply change.

6. Rent for student room in Cardiff(a)  H 0 : µ = 220 x − µ0 230 − 220  z= =  H 1 : µ ≠ 220 s n 35 40 Reject H0 if Z<-1.96 or if Z>1.96. =1.81 Thus we do not reject H0, i.e. the claim of £220 per month cannot be rejected.(b) s 35 C.I. for µ = x ± Z 0.025 * = 230 ± 1.96 * n 40 219.15 < µ < 240.85 Do not reject H0 since £220 is in the above interval.

7. A bank manager & her customers (a) 20  H 0 : p = 0.30 p= ˆ = 0.20  100  H 1 : p ≠ 0.30 0.30 * 0.70 σp = ˆ = 0.045 100 ˆ P − P0 0.20 − 0.30 Z= = = −2.22 < −1.96 σP ˆ 0.045 (b) Therefore, we reject at 95% level of confidence. Reject H0 if Z<-2.57. Thus do not reject H0 at 1% level of significance.

Thank You for Your Attention• To download this slide (PPT & PDF) visit www.pairach.com/teaching • Any further question? Please email to me@pairach.com

BS1501 tutorial 3_final

by Champ Pairach on Mar 11, 2012

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