8.14 A computer company claims that the batteries in its laptops last for 4 hour

Question

8.14 A computer company claims that the batteries in its laptops last for 4 hours on average. A consumer report firm gathered a sample
of 16 batteries and conducted tests on this claim. The sample mean was 3 hour 50 minutes and the sample standard deviation was 20 minutes.
Assume that the battery time is distributed as normal.

a.) Test if the average battery time is shorter than 4 hours at =0.05

b.)Construct a 95% confidence interval of the mean battery time.

c.) If you were to test H0: =240 minutes versus H1: 240 minutes, what would you conclude from your result in

part (b)?

d.) Suppose that a further study establishes that, in fact, the population mean is 4 hours. Did the test in part (c) make a correct decision?

If not, what type of error did it make?

Explanation / Answer

a.
Set Up Hypothesis
Null, H0: U=240
Alternate,average battery time is shorter than 4 hours H1: U<240
Test Statistic
Population Mean(U)=240
Sample X(Mean)=230
Standard Deviation(S.D)=20
Number (n)=16
we use Test Statistic (t) = x-U/(s.d/Sqrt(n))
to =230-240/(20/Sqrt(16))
to =-2
| to | =2
Critical Value
The Value of |t | with n-1 = 15 d.f is 1.753
We got |to| =2 & | t | =1.753
Make Decision
Hence Value of | to | > | t | and Here we Reject Ho
P-Value :Left Tail -Ha : ( P < -2 ) = 0.03197
Hence Value of P0.05 > 0.03197,Here we Reject Ho

We have evidence that average battery time is shorter than 4 hours

b.
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)=230
Standard deviation( sd )=20
Sample Size(n)=16
Confidence Interval = [ 230 ± t a/2 ( 20/ Sqrt ( 16) ) ]
= [ 230 - 2.131 * (5) , 230 + 2.131 * (5) ]
= [ 219.345,240.655 ]

c.
Failed to Reject Ho, as the reason 240 lies in the interval

d.
Yes,it makes the right decision

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