Lyme disease is spread in the northeastern United States by infected ticks. The ticks are infected mainly by feeding on mice, so more mice result in more infected ticks. The mouse population in turn rises and falls with the abundance of acorns, their favored food. Experimenters studied two similar forest areas in a year when the acorn crop failed. They added hundreds of thousands of acorns to one area to imitate an abundant acorn crop, while leaving the other area untouched. The next spring, 54 of the 72 mice trapped in the first area were in breeding condition, versus 10 of the 17 mice trapped in the second area. Required:
A 90% confidence interval for the difference between the proportions of mice ready to breed in good acorn years and bad acorn years is:

90% confidence interval is; ( - 0.0516, 0.3752 )

Step-by-step explanation:

Given the data in the question;  

n1 = 72, n2 = 17

P1 = 54 / 72 = 0.75

P2 = 10 / 17 = 0.5882

so

P_good = 0.75

P_bad = 0.5882

standard ERRROR will be;

SE = √[(0.75×(1-0.75)/72) + (0.5882×(1-0.5882)/17)]

SE = √( 0.002604 + 0.01424)

SE = 0.12978

given confidence interval = 90%

significance level a = (1 - 90/100) = 0.1, |Z( 0.1/2=0.05)| = 1.645   { from standard normal table}

so

93% CI is;

(0.75 - 0.5882) - 1.645×0.12978 <P_good - P_bad<  (0.75 - 0.5882) + 1.645×0.12978

⇒0.1618 - 0.2134 <P_good - P_bad<  0.1618 + 0.2134

⇒ - 0.0516  <P_good - P_bad< 0.3752

Therefore 90% confidence interval is; ( - 0.0516, 0.3752 )

coin flip

non-answer:

choosing

sorry if this isnt what you're looking for i dont really understand the question

see below

step-by-step explanation:

7.   there are 10 data points in the set.   the middle point is between 5 and 6.   the middle point in the top half is 8   and the middle point in the bottom half is 3. to find the iqr , take point 8 and subtract point 3

mid term   point 8 = 100   point 3 = 88

iqr = 100-88 = 12

final   point 8 = 93   point 3 = 78

iqr = 93 - 78 = 15

the final has the greatest iqr

6.   b - there is a high data value that causes the data set to be asymmetrical for the males: the data for males are high and asymmetrical .   there is an outlier for the males that skews the data

5.   the   iqr is a better measure of spread for the movies and basketball games.   since the basketball game has an outlier, we do not want to use standard deviation for measurement.  

3.   the exam median is much higher than the class median.   the median line is farther to the right on the exams

2.   there is no way to tell what the means are from the box and whisker plots.