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
