Abearing used in an automotive application is required to have a nominal inside diameter of 1.5 inches. a random sample of 25 bearings is selected and the average inside diameter of these bearings is 1.4975 inches. bearing diameter is known to be normally distributed with standard deviation inch. (a) test the hypotheses versus using the true mean hole diameter significantly different from 1.5 in. at alpha equals 0.01. (b) what is the p-value for the test in part (a)

a. You cannot conclude that the true mean of the inside diameter is different from 1.5 inches.

b. p-value: 0.992

Step-by-step explanation:

Hello!

To test the hypothesis that "The true mean of inside diameter is different from 1.5 inches."

Let's define the study variable:

X: "inside diameter of a bearing"

X~N(μ;)

With known standart derivation σ= 1 inch

Sample taken:

n= 25 bearings

X(bar)= 1.4975 inches

>Statistical hyphotesis for this problem

H₀: μ = 1.5

H₁: μ ≠ 1.5

>Level of significance

α: 0,01

>Since the variable has a normal distribution and the population standard desviation is known the mos accurate statistical to use is:

>The critical region for this test is two-tailed, that means you'll have two critical values.

This means, you'll reject the null hyphotesis if your is ≤ -2,58 or if it is ≥ 2,58

Now calculate the value by replacing the formula:

Since the calculated Z value is between the non rejection area, the test is not statistically significant. In other words, you cannot conclude that the true mean of the inside diameter is diferent from 1.5 inches.

b. To calculate the p-value, you have to look for the probability of the calculated Z-value

P(Z≤-0.0125) ≅ 0,496

The p-value above is one tailed, to have the p-value for this thest you need to simply multiply it by 2, so the p-value for the test is 0.992 (aprox)

the one that has a slope of 1/2 and is positive.

step-by-step explanation: 2 buckets 1 trip ; 4 buckets 2 trips and so on.

the answer i believe is 27 units.

step-by-step explanation: