Question

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.

asked Jun 15, 2021 182k views

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Tonja · Answer 1 · 2021-06-18T22:38:51+0000

Answer:

First step : State the null and the alternative hypothesis

The null hypothesis is
$H_o : \mu = 3.5 \ hours$

The alternative hypothesis is
$H_a : \mu > 3.5 \ hours$

Second Step : Calculate the test statistics

Generally the test statistics is mathematically represented as

$z = (\= x - \mu )/((\sigma )/(√(n) ) )$

=>
z = (3.75 - 3.5 )/((0.7 )/(√(45) ) )

=>
z = 2.11

Third Step : Obtain the p-value

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

P(Z > 2.11) = 0.0174

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

From the obtained value the
$p-value  <  \alpha$ 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

Explanation:

From the question we are told that

The population mean is
$\mu = 3.5 \ hours$

The sample size is n = 45

The standard deviation is
$\sigma = 0.7 \ hours$

The sample mean is
$\= x = 3.75$

The level of significance is
$\alpha = 0.05$