A phone manufacturer wants to compete in the touch screen phone market. He under

Question

A phone manufacturer wants to compete in the touch screen phone market. He understands that the lead product has a battery life of just 6 hours. The manufacturer claims that while the new touch phone is more expensive, its battery life is more than twice as long as that of the leading product. In order to test the claim, a researcher samples 42 units of the new phone and finds that the sample battery life averages 12.6 hours with a sample standard deviation of 1.7 hours. Use Table 2. a. Select the relevant null and the alternative hypotheses. a)H0: 12; HA: > 12 b) H0: = 12; HA: 12 c) H0: 12; HA: < 12 b. Compute the value of the appropriate test statistic. (Round intermediate calculations to 4 decimal places. Round your answer to 2 decimal places.) Test statistic c. Calculate the critical value to test the phone manufacturer’s claim at = 0.01. (Round your answer to 3 decimal places.) Critical value d-1 Approximate the p-value. a) 0.005 > p-value > 0.010 b) 0.025 > p-value > 0.050 c) 0.010 > p-value > 0.025 d-2. What is the conclusion? a) Do not reject H0; the battery life is not more than 12 hours b) Do not reject H0; the battery life is more than 12 hours c) Reject H0; the battery life is not more than 12 hours d) Reject H0; the battery life is more than 12 hours

Explanation / Answer

a)

Formulating the null and alternative hypotheses,              
              
Ho:   u   <=   12  
Ha:    u   >   12   [ANSWER, A]

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b)

Getting the test statistic, as              
              
X = sample mean =    12.6          
uo = hypothesized mean =    12          
n = sample size =    42          
s = standard deviation =    1.7          
              
Thus, t = (X - uo) * sqrt(n) / s =    2.287320246   [ANSWER]

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c)

Thus, getting the critical t,      

df = n - 1 =    41  

By table,

tcrit =    +   2.421 [ANSWER]

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d-1)

              
As we can see, this is a    right   tailed test.      
              
df = n - 1 =    41          

Hence, by table,

OPTION C: 0.010 < P < 0.025 [ANSWER]          

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d-2)

As P > 0.01, we do no reject Ho, so

OPTION A: a) Do not reject H0; the battery life is not more than 12 hours [ANSWER, A]

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