Four women have to cross a bridge safely under 60 minutes

A company is approached by a new warehouse management systems vendor to adopt a new system. The vendor claims that the warehouse

Ludmilka [50]

Answer:

a

The null hypothesis is $H_o : \mu = 180$

The alternative hypothesis is $H_a : \mu < 180$

b

The decision rule is fail to reject the null hypothesis

The conclusion

There no sufficient evidence to support the vendor claim that the warehouse management system reduces the average pick, pack, and ship time to below 3 minutes(180 seconds) per order through bin location and routing optimization

Step-by-step explanation:

From the question we are told that

The population mean is $\mu = 3 \ minutes = 180 \ seconds$

The sample size is n = 25

The excel sheet is show on the first uploaded image

The level of significance is $\alpha = 0.05$

The null hypothesis is $H_o : \mu = 180$

The alternative hypothesis is $H_a : \mu < 180$

Generally the sample mean is mathematically represented as

$\= x = \frac{156 + 136 + \cdots + 181}{25}$

=> $\= x = 173.2$

Generally the standard deviation is mathematically represented as

$\sigma = \sqrt{\frac{\sum (x_i - \= x )^2 }{n} }$

=> $\sigma = \sqrt{\frac{ (156 - 173.2 )^2 + (136 - 173.2 )^2 + \cdots + (181- 173.2 )^2 }{25} }$

=> $\sigma = 24.261$

Generally the test statistics is mathematically represented as

$t = \frac{\= x - \mu }{\frac{\sigma}{\sqrt{n} } }$

=> $t = \frac{173.2 - 180 }{\frac{24.261}{\sqrt{25} } }$

=> $z = -1.40$

Generally p-value is mathematically represented as

$p-value = P(Z < z)$

=> $p-value = P(Z < -1.40)$

From the z-table

$P(Z < -1.40) = 0.080757$

=> $p-value = 0.080757$

From the obtained value we that $p-value > \alpha$

The decision rule is fail to reject the null hypothesis

The conclusion

There no sufficient evidence to support the vendor claim that the warehouse management system reduces the average pick, pack, and ship time to below 3 minutes(180 seconds) per order through bin location and routing optimization