In 2002, the mean expenditure for auto insurance in a certain state was $806. An insurance salesperson in this state believes that the mean expenditure for auto insurance is less today. She obtains a simple random sample of 32 auto insurance policies and determines the mean expenditure to be $781 with a standard deviation of $39.13. Is there enough evidence to support the claim that the mean expenditure for auto insurance is less than the 2002 amount at the α = 0.05 level of significance?

No there is not  enough evidence to support the claim that the mean expenditure for auto insurance is less than the 2002 amount at the α = 0.05 level of significance

Step-by-step explanation:

Sample Mean = μ1 = $806

Sample Mean = μ2 = $ 781

Standard Deviation= S= σ =39.13

n= 32

Confidence Interval = 95 %

α= 0.05

z∝=± 1.96

We state the null and alternative hypotheses as

H0: μ1 = $806  and Ha: μ1 ≠ $806  two sided tail test

z= μ1 -μ2/σ/√n

z= 806-781/ 39.13/√32

z= 806-781/ 39.13/5.6568

z=806-781/ 6.92

z= 25/6.92

z= 3.613

Z> z∝

3.613 > ± 1.96

No there is not  enough evidence to support the claim that the mean expenditure for auto insurance is less than the 2002 amount at the α = 0.05 level of significance

$20.01

step-by-step explanation:

the sale price is (1.00 - 0.77)($87), or 0.23($87), or $20.01.

55.44 inches

step-by-step explanation:

part a. assuming a perfectly linear relationship, we can find the slope from the first two data points.

slope = m = (change in rainfall)/(change in years)

= (61.80 -62.33)/(2005 -2004) = -0.53/1 = -0.53

then the point-slope form of the equation of the line can be written as

y = m(x -h) +k . . . . . m = -0.53, (h, k) = (0, 62.33)

y = -0.53x +62.33 . . . x = years after 2004

part b. in 2017, x = 2017 -2004 = 13. then the predicted rainfall is

y = -0.53·13 +62.33 = 55.44 . . inches

the predicted rainfall in 2017 is 55.44 in.