In the past, the mean running time for a certain type of flashlight battery has been 8.6 hours. The manufacturer has introduced a change in the production method and wants to perform a hypothesis test to determine whether the mean running time has increased as a result.
The hypotheses are:
H0: μ = 8.6 hours
Ha: μ > 8.6 hours
where μ is the mean running time of the new batteries.
Explain the meaning of a Type II error. A. The manufacturer will decide the mean battery life is greater than 8.3 hours when in fact it is greater than 8.3 hours. B. The manufacturer will decide the mean battery life is less than 8.3 hours when in fact it is greater than 8.3 hours. C. The manufacturer will decide the mean battery life is 8.3 hours when in fact it is 8.3 hours. D. The manufacturer will decide the mean battery life is 8.3 hours when in fact it is greater than 8.3 hours. E. The manufacturer will decide the mean battery life is greater than 8.3 hours when in fact it is 8.3 hours.

H0: μ = 8.3 hours

Ha: μ > 8.3 hours

We need to remember that a type of error II if the error associated when we NO reject the null hypothesis when actually the alternative hypothesis is true. And is also known as false negative. The best answer for this case would be:

D. The manufacturer will decide the mean battery life is 8.3 hours when in fact it is greater than 8.3 hours.

Since we are FAILING to reject the null hypothesis when the true mean is actually higher than the value specified

Step-by-step explanation:

For this case we want to determine if mean running time has increased as a result of the new change so then the system of hypothesis are given by:

Note: we assume that the value to check in order to be consistent with the options is 8.3

H0: μ = 8.3 hours

Ha: μ > 8.3 hours

We need to remember that a type of error II if the error associated when we NO reject the null hypothesis when actually the alternative hypothesis is true. And is also known as false negative. The best answer for this case would be:

D. The manufacturer will decide the mean battery life is 8.3 hours when in fact it is greater than 8.3 hours.

Since we are FAILING to reject the null hypothesis when the true mean is actually higher than the value specified

intensity of the vector is v= √37 ≈ 6.08 units and make angle ∡α ≈ 9.46° with east direction

step-by-step explanation:

required vector is consists of the two componentsvx= 2+4=6 units   and vy= 1 unit and vx ⊥ vywe will use pythagorian theorem to find intensity of the vector vv∧2 = vx∧2 + vy∧2 => v = √vx∧2 + vy∧2 = √6∧2 + 1∧2 = √36+1 = √37 ≈ 6.08 unitsthe angle ∡α between vector and east direction we wil find with tanαtanα = 1/6 => α = arc tanα = arc 1/6 => α ≈ 9.46°

hope this