University of Alberta

• STAT 235

• SOLUTIONS TO HOMEWORK ASSIGNMENT 4 8.1.4 Assumptions: a) Simple random sample? Yes b) n > 30 OR population is normal? Since n = 16, perhaps we assume that the population is normal. But there is not enough information to check this assumption. Since σ is unknown, we use a t-distribution with df = n – 1 = 16 – 1 = 15. The two-sided 99% confidence interval for µ (with t0.005, 15 = 2.947) is x ± tα/2, df s n        0.058 1.053 (2.947) 16         1.053 0.0427  (1.010, 1.096) Interpretation: It is estimated with 95% confidence that the average sugar packet weight is between 1.010 kilograms and 1.096 kilograms. Since the interval contains the value 1.025, it is plausible that the average sugar packet weight is 1.025 kilograms. (The conclusion may be invalid due to the failure of the normality assumption.) 8.1.8 With t0.005, n – 1 ≈ 3, a sufficient sample size can be estimated as n ≥ 2 2 /2, 1 0 3*0.15 4* 4* 20.250 21 0.2 n t s L                   df = 21 – 1 = 20  t0.005, 20 = 2.845 n ≥ 2 2 /2, 1 0 2.845*0.15 4* 4* 18.212 19 0.2 n t s L                  Notice that t0.005, n – 1 can also be estimated as 4 (more conservative approach correct, even for relatively small sample sizes; see the t-table). In this case, n ≥ 2 2 /2, 1 0 4*0.15 4* 4* 36 0.2 n t s L                  df = 36 – 1 = 35 ≈ 30  t0.005, 30 = 2.750 n ≥ 2 2 /2, 1 0 2.750*0.15 4* 4* 17.016 18