A computer company claims its laptop batteries averages more than 3.5 hours ofnuse per charge. A sample of 45 batteries last average of 3.72 hours. Assumes the sample and standard deviation of 0.7 hours. What are rhe correct steps to follow to figure out if you can verify the company's alpha of 0.05?

Answer:

First step : State the null and the alternative hypothesis

   The null hypothesis is  [tex]H_o : \mu = 3.5 \ hours[/tex]

    The alternative hypothesis is  [tex]H_a : \mu > 3.5 \ hours[/tex]

Second Step : Calculate the test statistics

   Generally the test statistics is mathematically represented as

             [tex]z = \frac{\= x - \mu }{\frac{\sigma }{\sqrt{n} } }[/tex]

=>           [tex]z = \frac{3.75 - 3.5 }{\frac{0.7 }{\sqrt{45} } }[/tex]

=>           [tex]z  =  2.11[/tex]

Third Step  : Obtain the p-value

Generally from the z-table the probability of  z  =  2.11 is

    [tex] P(Z  >  2.11) =  0.0174[/tex]

Fourth Step : Compare the p-value with the level of significance and state the decision rule and the conclusion

From the obtained value  the [tex]p-value  <  \alpha[/tex] hence

 The decision rule is  

The null hypothesis is rejected

 The conclusion is

There is sufficient evidence to conclude that average of the laptop batteries is more than 3.5 hours of use per charge

Step-by-step explanation:

From the question we are told that

  The population mean is [tex]\mu = 3.5 \ hours[/tex]

   The sample size is  n =  45

   The standard deviation is  [tex]\sigma = 0.7 \ hours[/tex]

   The sample mean is  [tex]\= x = 3.75[/tex]

    The level of significance is  [tex]\alpha = 0.05[/tex]

First step : State the null and the alternative hypothesis

   The null hypothesis is  [tex]H_o : \mu = 3.5 \ hours[/tex]

    The alternative hypothesis is  [tex]H_a : \mu > 3.5 \ hours[/tex]

Second Step : Calculate the test statistics

   Generally the test statistics is mathematically represented as

             [tex]z = \frac{\= x - \mu }{\frac{\sigma }{\sqrt{n} } }[/tex]

=>           [tex]z = \frac{3.75 - 3.5 }{\frac{0.7 }{\sqrt{45} } }[/tex]

=>           [tex]z  =  2.11[/tex]

Third Step  : Obtain the p-value

Generally from the z-table the probability of  z  =  2.11 is

    [tex] P(Z  >  2.11) =  0.0174[/tex]

Fourth Step : Compare the p-value with the level of significance and state the decision rule and the conclusion

From the obtained value  the [tex]p-value  <  \alpha[/tex] hence

 The decision rule is  

The null hypothesis is rejected

 The conclusion is

There is sufficient evidence to conclude that average of the laptop batteries is more than 3.5 hours of use per charge

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