10 100 c. 1000 d. 10000 \" 14. Data set A has normal distribution; A ~ N(, 2) wh

Question

10 100 c. 1000 d. 10000 " 14. Data set A has normal distribution; A ~ N(, 2) where and 2 are unknown. A sample of 30 S/TN= 0.137 out bound 91.11 · subjects are selected from this data set with sample mean = 92 and sample standard deviation s -0.75, 90% confidence interval for population mean () would be: a. 90.01-95.45 b. 93.34-93.28 c 91.77-92.23 SEM, 91.10-96.01 Up pti bourd 9323 15. Assume that 50% of babies born are boys. In a small city, on a particular day, 8 babies were born and 6 of them were boys. On the same day in a large state, 800 babies were born and 600 of them were boys. Which statement do you agree with? a. What occurred in the city was more unusual/unexpected than what occurred in the state. b. What occurred in the state was more unusual/unexpected than what occurred in the city c· The occurrences in both locales were equally unusual or unexpected d. Nothing seems unusual in either locale. 6. We test the null hypothesis that says population mean () is 110 (H0: = 110). We draw a ) of 75, and the sample mean O) is 125 with sample standard deviation (s) of 10, The sample (n value of test statistic [Z or i] upon which we may (or may not) reject Ho would be equal to a. 10.65 h. 15 960 5-110

Explanation / Answer

#14.

#15.

what occured in the state was more unusual/unexpected than what occurred in the city.

This is because,

p = 0.5

For city if we calculate the test statistics, it is z = 1.4142

and for state is it 14.1421

Obviously for the state we have p-value rejection of null hypothesis. Hence it is unusual proportion of boys born

CI for 90% n 30 mean 92 z-value of 90% CI 1.6449 std. dev. 0.75 SE = std.dev./sqrt(n) 0.13693 ME = z*SE 0.22523 Lower Limit = Mean - ME 91.77477 Upper Limit = Mean + ME 92.22523 90% CI (91.7748 , 92.2252 )

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