Read each question carefully and Show all relevant work. You must show enough wo

Question

Read each question carefully and Show all relevant work. You must show enough work to arrive at and justify an answer. Answers with little or no supporting work will receive little credits. Correct answers with incorrect supporting work will receive no credit, and will (text not clear). A manufacturer claims that a new beand of airconditioning unit uses only 6.5 kilowatts of electricity per day. A consumer agency believes the true figure to be higher and rans a test on a random sample of size 50. If the sample mean is 7.0 kilowatts, with a standard deviation of 1.4, should the manufacturers claim be rejected at a significance level of 1% What would the null and alternative hypothesis be if you wished to test if the true electricity usage was more than the advertised usage Given that you are testing at a 90% level of significance (alpha = 0.01), what would be the value of the test statistic that would be used to establish the rejection region Perform a hypothesis test to determine if the electricity usage claim is true.

Explanation / Answer

t-test For Single Mean
Set Up Hypothesis
Null, H0: U=6.5
Alternate, H1: U>6.5
Test Statistic
Population Mean(U)=6.5
Sample X(Mean)=7
Standard Deviation(S.D)=1.4
Number (n)=50
we use Test Statistic (t) = x-U/(s.d/Sqrt(n))
to =7-6.5/(1.4/Sqrt(50))
to =2.525
| to | =2.525
Critical Value
The Value of |t | with n-1 = 49 d.f is 2.405
We got |to| =2.525 & | t | =2.405
Make Decision
Hence Value of | to | > | t | and Here we Reject Ho
P-Value :Right Tail - Ha : ( P > 2.5254 ) = 0.00742
Hence Value of P0.01 > 0.00742,Here we Reject Ho

ANSWERS
a.
H0: U=6.5 ,H1: U>6.5

b.
Critical Value 2.405

c.
the true elecricity usage was more than the advatersized value

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