Anew design for the braking system on a certain type of car has been proposed. for the current system, the true average braking distance at 40 mph under specified conditions is known to be 120 ft. it is proposed that the new design be implemented only if sample data strongly indicates a reduction in true average braking for the new design. define the parameter of interest and state the relevant hypothesis. suppose braking distance for the new system is normally distributed with sigma = 10. let x denote the sample average braking distance for a random sample of 36 observations. which of the following three rejection regions is appropriate : what is the significance level for the appropriate region of part(b)? how would you change the region to obtain a test with alpha = 0.001? what is the probability that the new design is not implemented when its true average braking distance is actually 115 ft and the appropriate region from part (b) is used? let z = (x - 120)/ (sigma / n), what is the significance level for the region {z : z le -2.33}? for the region {z : z le -2.88}?

Step-by-step explanation:

Hello!

a. The study variable is X: "breaking distance using the new design"(ft)

X~N(μ;σ²)

The parameter of interest is μ: The population mean of the breaking distance using the new design.

The hypothesis states that the new design reduces the population mean of the braking distance. Symbolically:

H₀: μ≥120

H₁: μ<120

Unfortunately, you didn't copy the choices for the critical regions. With this type of hypothesis, you have a one-tailed critical region and will reject the null hypothesis to small values of the statistic. I've looked for the critical regions using the most common significance levels and the one in part b.

Significance level α: 10%

Critical value

Significance level α: 5%

Significance level α: 1%

b. Significance level α: 0.1%

The significance level is stated by the investigator before making the test, what changes is the rejection region, with a lower level of significance you get a smaller rejection region and is more likely to support a false null hypothesis. On the other hand, the smaller the level of significance, the bigger the power of the test so if you end up rejecting the null hypothesis you will have more certainty of having made a correct decision.

c. The null hypothesis states that the population mean of the braking distance is 120ft or more and you conduct the test supposing this hypothesis is true, that's why you can use the number in this hypothesis when calculating the statistic value.

If later it is found out that the population mean is 115 ft, that means your null hypothesis is FALSE.

And if the condition for implementing the new system is that the braking distance is reduced, since it wasn't applied you can assume that the null hypothesis WASN'T REJECTED.

So, in this case, it is a Type II error situation, You didn't reject the null hypothesis but it was false and its associated probability is β.

d.

{z : z le -2.33} significance level is 0.0099 ≅ 0.01

{z : z le -2.88} significance level is 0.00199 ≅ 0.02

I hope it helps!

the results from a pre-test for students for the year 2000 and the year 2010 are illustrated in the box plot. what do these results tell us about how students performed on the 29 question pre-test for the two years?

if we compare only the lowest and highest scores between the two years, we might conclude that the students in 2010 did better than the students in 2010. this conclusion seems to follow since the lowest score of 8 in 2010 is greater in value than the lowest score of 6 in 2000. also, the highest score of 28 in 2010 is greater in value than the highest score of 27 in 2000.

but the box portion of the illustration gives us more detailed information. the middle bar in each box shows us that the median score of 20 in 2000 is greater in value than the median score of 17 in 2010. further, we note that the box and whiskers divide the illustration into four pieces. each of these four pieces represents the same portion of students. so, the upper half of the students in 2000 scored in the same score range as the upper one-fourth of the students in 2010, see the illustration at a score of 20.

by considering the upper one-fourth, upper half, and upper three-fourths instead of just the lowest and highest scores, we would conclude that the students as a whole did much better in 2000 than in 2010. we would conclude that as a whole the students in 2010 are less prepared than the students in 2000.

in this section, we discuss box-and-whisker plots and the five key values used in constructing a box-and-whisker plot. the key values are called a five-number summary, which consists of the minimum, first quartile, median, third quartile, and maximum.

step-by-step explanation:

u = -(6/7)w

step-by-step explanation:

place all values with 'u' on one side of the equation.

4u + 8w = - 3u + 2w

4u + 3u = 2w - 8w

further simplify the equation by solving the addition and subtraction.

7u = - 6w

make 'u' the independent variable by placing its co-efficient on the right-hand side.

u = - (6/7)w