Before agreeing to purchase a large order of polyethylene sheaths for a particular type of high-pressure oil-filled submarine power cable, a company wants to see conclusive evidence that the true standard deviation of sheath thickness is less than 0.05 mm. What hypotheses should be tested, and why?The appropriate hypotheses areH0: σ0.05 mmversusHa: σ0.05 mm. With this formulation, the burden of proof is on the data to show that the requirementbeen met. In this context, what are the type I and type II errors?In this context, the type I error occurs if wea shipment that should have been . A type II error occurs if we a shipment that should have been. Need Help? Read It Talk to a Tutor

Null and alternative hypothesis:

The alternative hypothesis is the one that needs evidence to be supported, while the null hypothesis is the one that can be nullified (reject).

Only if there is enough evidence that thickness is less than 0.05 the null hypothesis will be rejected and the alternative hypothesis claim supported.

A Type I error happens when a true null hypothesis is rejected. In this case we will be purchase a order that is not fulfilling the thickness required.

A Type II error happens when a false null hypothesis is failed to be rejected. In this case, the order has a thickness significantly smaller than 0.05, but the sample gives no enough evidence and the order will not be purchased.

Step-by-step explanation:

A hypothesis test to see conclusive evidence that the true standard deviation of sheath thickness is less than 0.05 mm will have the following hypothesis:

The alternative hypothesis Ha will state that the true mean is significantly smaller than 0.05, while the null hypothesis H0 will state the opposite: that the true mean is not significantly smaller than 0.05.

The alternative hypothesis is the one that needs evidence to be supported, while the null hypothesis is the one that can be nullified (reject).

Only if there is enough evidence that thickness is less than 0.05 the null hypothesis will be rejected and the alternative hypothesis claim supported.

whole numbers.

step-by-step explanation:

we are given the average monthly rainfall (in millimeters) model r(m) for the years 1900-2009 as a function of the month number m.

as we can see m is taken for the number of months for years 1900-2009.

we know that number of months can't be a negative number or decimal number or a fraction.

so, the number of months could be only positive whole numbers.

therefore, whole numbers is more appropriate for the domain of r.