Consider the distribution of 5years survival for individuals

•Consider the distribution of 5-years survival for individuals under 40 who have been diagnosed with lung cancer. In a random selected sample of 52 patients, only 6 survive 5 years. Estimate with 95% confidence level the proportion of patients who survive more than 5 years.if more than 12% of households participate in it.

please by using Minitab

Solution

sample proportion = 6/52 = 0.1153846...s.d = sqrt [ 0.1153846*(1- 0.1153846) / 52 ] = 0.04430467..

basic stat>1 sample t test>summarised data..

Test of mu = 0.12 vs > 0.12

95% Lower
N Mean StDev SE Mean Bound T P
52 0.11538 0.04430 0.00614 0.10509 -0.75 0.772..

so, popltn proportion is 0.12...