A manufacturer produces piston rings for an automobile engine. It is known that ring diameter is normally distributed with millimeters. A random sample of 15 rings has a mean diameter of x Overscript bar EndScripts equals 74.127. Construct a 99% two-sided confidence interval on the true mean piston diameter and a 95% lower confidence bound on the true mean piston diameter. (b) Calculate the 95% one-sided lower confidence interval on the true mean piston diameter.

(a) 74.12633 mm ≤ μ ≤ 74.12767 mm

(b) μ ≥ 74.12649 mm

Step-by-step explanation:

Here we have;

Sample count, n = 15

Mean, = 74.127 mm

Standard deviation, σ = 0.001 mm

(a) The confidence interval, CI is given as follows;

At 99%, z = ±2.575829

Therefore, the confidence interval is;

74.12633 mm ≤ μ ≤ 74.12767 mm

(b) The lower confidence at 95% is given by;

z at 95% = 1.959964

or μ ≥ 74.12649 mm.

a 4/6 = 9/12   not proportional

step-by-step explanation:

to find which are not proportional, we can use cross products

if it is true, they are proportional.   if it is not true, they are not proportional.

a 4/6 = 9/12

4* 12 = 6*9

48 = 54

not this is not proportional  

b 2/3 = 12/18  

2* 18 = 3 * 12

36 = 36

this is proportional

c 5/4 = 10/8  

5*8 = 4*10

40 = 40

this is proportional

d 7/5 = 35/25

7*25 = 5 * 35

175 = 175

is proportional