A survey conducted by the American Automobile Association showed that a family o

Question

A survey conducted by the American Automobile Association showed that a family of four spends an average of $215.60 per day while on vacation. Suppose a sample of 64 families of four vacationing at Niagara Falls resulted in a sample mean of $252.45 per day and a sample standard deviation of $77.50. Develop a 95% confidence interval estimate of the mean amount spent per day by a family of four visiting Niagara Falls (to 2 decimals). (,) 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?

Explanation / Answer

a.

Confidence Interval
CI = x ± t a/2 * (sd/ Sqrt(n))
Where,
    x    = Mean
    sd   = Standard Deviation
    a    = 1 - (Confidence Level/100)
    ta/2 = t-table value
    CI   = Confidence Interval
Mean(x)=252.45
Standard deviation( sd )=77.5
Sample Size(n)=64
Confidence Interval = [ 252.45 ± t a/2 ( 77.5/ Sqrt ( 64) ) ]
= [ 252.45 - 1.998 * (9.688) , 252.45 + 1.998 * (9.688) ]
= [ 233.094,271.806 ]

b.

Null, H0: U=21.6
Alternate, H1: U!=21.6
Test Statistic
Population Mean(U)=21.6
Sample X(Mean)=252.45
Standard Deviation(S.D)=77.5
Number (n)=64
we use Test Statistic (t) = x-U/(s.d/Sqrt(n))
to =252.45-21.6/(77.5/Sqrt(64))
to =23.83
| to | =23.83
Critical Value
The Value of |t | with n-1 = 63 d.f is 1.998
We got |to| =23.83 & | t | =1.998
Make Decision
Hence Value of | to | > | t | and Here we Reject Ho
P-Value :Two Tailed ( double the one tail ) -Ha : ( P != 23.8297 ) = 0
Hence Value of P0.05 > 0,Here we Reject Ho

Yes, it is differ

Confidence Interval
CI = x ± t a/2 * (sd/ Sqrt(n))
Where,
    x    = Mean
    sd   = Standard Deviation
    a    = 1 - (Confidence Level/100)
    ta/2 = t-table value
    CI   = Confidence Interval
Mean(x)=252.45
Standard deviation( sd )=77.5
Sample Size(n)=64
Confidence Interval = [ 252.45 ± t a/2 ( 77.5/ Sqrt ( 64) ) ]
= [ 252.45 - 1.998 * (9.688) , 252.45 + 1.998 * (9.688) ]
= [ 233.094,271.806 ]

b.

Null, H0: U=21.6
Alternate, H1: U!=21.6
Test Statistic
Population Mean(U)=21.6
Sample X(Mean)=252.45
Standard Deviation(S.D)=77.5
Number (n)=64
we use Test Statistic (t) = x-U/(s.d/Sqrt(n))
to =252.45-21.6/(77.5/Sqrt(64))
to =23.83
| to | =23.83
Critical Value
The Value of |t | with n-1 = 63 d.f is 1.998
We got |to| =23.83 & | t | =1.998
Make Decision
Hence Value of | to | > | t | and Here we Reject Ho
P-Value :Two Tailed ( double the one tail ) -Ha : ( P != 23.8297 ) = 0
Hence Value of P0.05 > 0,Here we Reject Ho

Yes, it is differ

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