Question

smple of s mconded dn the United States follows a normal distribution. A random sample of 64 recorded deaths in the United of 71 8 years with a standard deviation of S.9 years. (20pes) 6 Assume that the life span in follows a States during the past year showed an average life span (a) (10pts) Construct intervail of the mcan life span narrower or wider? Explain your reasoning. bi (00pe) Does ycars? Use a 005 level of significance and follow the hypothesis test procedures a95% confidence interval of the mean life span, would a 90% confidence the data collected seem to indicate that the mean life span today is greater than 70

Answer 1

Solution: Sample size = n = 64 Sample mean =x = 71.8 Sample Standard Deviation =s 8.9 Degrees of freedom =df = n-1 64-1 = 63 Part-A: For 95% confidence interval, α 1-0.95 0.05 a/2 0.05/2 0.025 From t-distribution table, the value of t having an area of (a/2 = 0.025) in its upper tail and for (df-63) is given by = t0.025 = 1.998 Margin of Error: MoE = t0.025X3 8.9 MoE1.998X 64 MoE2.223 95% confidence interval for mean life span: Lower Limit of 95% confidence interval 71.8-2.223 Upper Limit of 95% confidence interval-71.8+ 2.223 69.577 74.023 For a 90% confidence interval, the margin of error will be smaller and hence, the width of the confidence interval will be narrower compared to a 95% confidence interval

Part-B: Hai μ 70 H1: μ> 70 Test Statistic: t= (718-70)/(89/ 「64) t = 1.618 From t-distribution table, area under the curve to the right of (t = 1.618) and for (df = 63, is given by: 0.0553 Hence, p-value 0.0553 Since p-value >α, we cannot reject the null hypothesis Conclusion: Fail to reject Ho. There is not sufficient evidence to conclude that the mean life span today is greater than 70 years