A survey conducted by the American Automobile Association (AAA) showed that a family of four spends an average of per day while on vacation. Suppose a sample of families of four vacationing at Niagara Falls resulted in a sample mean of per day and a sample standard deviation of . a. Develop a confidence interval estimate of the mean amount spent per day by a family of four visiting Niagara Falls (to decimals). $ to $ b. Based on the confidence interval from part (a), does it appear that the population mean amount spent per day by families visiting Niagara Falls differs from the mean reported by the American Automobile Association

(233.8436 ; 271.0564)

Yes

Step-by-step explanation:

Given :

Sample mean, xbar = 252.45

Sample standard deviation, s = 74.50

Sample size, n = 64

α = 0.05

The confidence interval :

Mean ± margin of error

Margin of Error = tα/2 * s/√n

df = n - 1 = 64 - 1 = 63

t(α/2, df = 63) = 1.998

Margin of Error = 1.998 * 74.50/√64

Margin of Error = 18.6064

The confidence interval :

252.45 ± 18.6064

(233.8436 ; 271.0564)

Comparing the confidence interval value and the mean value reported by the American Automobile Association ;it can be concluded that the mean reported by the American Automobile Association differ from the mean spent at Niagara Fall as 215.60 falls below the confidence interval.

hello :

by applying the distributive property :

6 × 4.3 = 6 × ( 4 + 0.3 )

            =   (6 ×   4) + (6 ×   0.3)

            = 24 + 1.8

6 × 4.3 = 25.8

the answer will be b x² +x-1

step-by-step explanation: