Question

m ure of International developmental progrebirth mortality rates are analyzed annually. For comtes reported the following birth mortality rates given at the end of the problem 26 15 75 66 18 52 125 5 6 58 119 98 32 5 12 13 99 6 22 20 88 34 103 (1) What would be an appropriate point estimate for the true birth mortality rate for these 25 countrie Defend your position. The appropriat point estimate the median because it would be stored Lould be the median o' (b) Identify the value of the point estimate chosen in part (a). The median for porta vanld be 30.. (c) Estimate the true birth mortality rate for these countries with a 95% level of confidence (d) Were you able to construct an interval with a 95% level of confidence? If not, why not?

Answer 1

Solo Giren data, 198 26 15 75 66 1852 19 13 99 6 1955 119 22 20 88 34103 32 n =23 a you can use any measure of Central tendency to depresent a point estimate for bioth modtality rate. I am here using Sample mean as a point estimate. Since Sample mean (x) which incorporates all observations, by The value of the point estimate X: 26 +15+75 +66 + 18 +58 +125 +5+6 +58 + 119 +98 + 32 +5+12+13 +99+6 + 2a + 20 + 88+ 34+103 23

B: 1097 1091 23 X = 47,69565. 9 95% Confidence Interval. Since 0:23 < 30 we can use t. distribution to obtain 95% Confidence interval. ne x = tala u Where tala as obtained from the table (t table with n-1 degrees of free dom) 5 95 the Sample Standard deviation - n Sample Size = 23 From t table ta/2=3.014 N -1

S: Sample Standard deviation N the Number of Observations X; - the observed Values of a sample item X: the mean value of the observation 5. (26-47.69365)+(15-47.69565)+(15-41-69565 | +(66-47-6956939+(18-47.695659+(52-47.69565 +(125-47.69565)+(5-47.69565)*+ (6-47.69565)= (58–47-69565) + (19-47.69565) + (98-47.69565- (32-47-6956999+ (5-47.69565° +62-47.69565) + C13 - 47.645699 +(99-47-69565)*+(6-47.69565) (22-47-69565)2 + (20-47.69565j9 +(88-47.69565 +(34-47.69565j9+(103-47.69565)2 머 3, 40,44911 o'95. Confidence interval is,

X. Mean t: t-value from table 9.704. si Standard deviation ne number of obsesvations. 47.69565 -2,704 x 40.44911 va3 47.69565 +9704 x 40.44911 123 (47.69565 -2.704 x 4044911 4.7958 ur. 69565+2-704 x u 4.7958 > (47.69565 - 3.704 x 8.4342, 47.69565 +2.704 x 8.4342) (24.88951, 70.50179) dy Since I used Sample mean as point estimatoo, I Can Construct 95% Confidence Potesval using t distribution