Question

Compute in excel

Text Question 8.3.8 (Confidence Interval for a Mean, not a proportion) (adapted) The Table below contains the percentage of women receiving prenatal care in 2009 for a sample of countries. (i) Find the 90% confidence interval for the average percentage of woman receiving prenatal care in 2009 using the normal distribution. (Express as percentages with two digits, as shown in the data.) (ii) Find the 90% confidence interval for the average percentage of woman receiving prenatal care in 2009 using the t-distribution. (Express as percentages with two digits, as shown in the data.) (iii) Are you answers similar? Why or why not? Note: You’ll find the data below reproduced in the Excel answer template. 70.08 72.73 74.52 75.79 76.28 76.28 76.65 80.34 80.60 81.90 86.30 87.70 87.76 88.40 90.70 91.50 91.80 92.10 92.20 92.41 92.47 93.00 93.20 93.40 93.63 93.68 93.80 94.30 94.51 95.00 95.80 95.80 96.23 96.24 97.30 97.90 97.95 98.20 99.00 99.00 99.10 99.10 100.0 100.0 100.00 100.00 100.00 (b) Text Question 8.2.5 (Confidence Interval for a Proportion) In 2013, the Gallup poll asked 1039 U.S. adults if they agreed that a conspiracy was behind the assassination of President Kennedy. And found that 634 did so. Using a 98% confidence interval, estimate the proportion of Americans who believe in this conspiracy. Express both lower and upper bounds as a percentage, rounding to 2 digits, e.g., 57.29, 65.34.

Answer 1

8.3.8)

from the given data use excel to get following value

sample size =n=47

sample mean =m=90.95

sample standard deviation=S=8.48

i)

90% confidence interval using normal distribution is given by

So interval is ( 88.92,92.98)

ii)

90% confidence interval using t distribution here df=n-1 =47-1 =46

Hence interval is (88.87,93.03)

Both confidence intervals are not exactly similar but approx both are same ,differences are basically due to normal distribution more concentrated towards its mean while t distribution have longer tail

8.2.5

given that n=1039

x=634

so

sample proportion =P=x/n =634/1039 =0.6102

then 98% confidence interval is given by

Hence interval is (0.575,0.6454)

Hence in % interval is (57.50,64.54)