Many older homes have electrical systems that use fuses rather than circuit breakers. A manufacturer of 40-amp fuses wants to make sure that the mean amperage at which its fuses burn out is in fact 40. If the mean amperage is lower than 40, customers will complain because the fuses require replacement too often. If the mean amperage is higher than 40, the manufacturer might be liable for damage to an electrical system due to fuse malfunction. To verify the amperage of the fuses, a sample of fuses is to be selected and inspected. If a hypothesis test were to be performed on the resulting data, what null and alternative hypotheses would be of interest to the manufacturer? Describe type I and type II errors in the context.

Type I error: Concluding μ ≠ 40, when in fact μ = 40.

Type II error: Concluding μ = 40, when in fact μ ≠ 40.

Step-by-step explanation:

In this case we need to determine whether the mean amperage at which the 40-amp fuses burn out is 40.

The hypothesis to test this can be defined as follows:

H₀: The mean amperage at which the 40-amp fuses burn out is 40, i.e. μ = 40.

Hₐ: The mean amperage at which the 40-amp fuses burn out is different from 40, i.e. μ ≠ 40.

A type I error occurs when we discard a true null hypothesis (H₀) and a type II error is made when we fail to discard a false null hypothesis (H₀).

In this context, a type I error will be committed if we conclude that the mean amperage at which the 40-amp fuses burn out is different from 40, when in fact it is 40.

And a type II error will be committed if we conclude that the mean amperage at which the 40-amp fuses burn out is 40, when in fact it is different from 40.

answer: 457.14

step-by-step explanation:

205 - 45 = 160

160 / .35 = 457.142857

let x and 400 - x be the two numbers.  

==> 0.80x + 0.85(400 - x) = 400 - 68 from the statement of the problem.  

< ==> 0.80x + 0.85(400 - x) = 400 - 68  

==> 340 - 0.05x = 332  

< ==> -0.05x = -8 ==> x = 160, the first number ==> 400 - x = 240, the second number.