A rivet is to be inserted into a hole. A random sample of parts is selected and the hole diameter is measured. The sample standard deviation is millimeters. Is there strong evidence to indicate that the standard deviation of hole diameter exceeds 0.01 millimeters

This question is incomplete, the complete question is;

A rivet is to be inserted into a hole. A random sample of n = 15 parts is selected and the hole diameter is measured. The sample standard deviation is s = 0.008 millimeters. Is there strong evidence to indicate that the standard deviation of hole diameter exceeds 0.01 millimeters, Use ∝ = 0.01

since our test statistics ( 8.96 ) is less than the critical value ( 29.141 ),

there is no strong evidence to indicate that the standard deviation of hole diameter exceeds 0.01 millimeters.

Step-by-step explanation:

Given that;

H₀ : σ = 0.01

H₁ : σ > 0.01

The test statistics is;

X² = ( n-1 )s² / σ²  = ( 15 - 1)(0.008)² / (0.01)²

= 0.000896 / 0.0001

= 8.96

degree of freedom

df = n - 1 = 15 - 1 = 14

For right tailed test, critical value at 0.01 level of significance is 29.141.

Now since our test statistics ( 8.96 ) is less than the critical value ( 29.141 ),

there is no strong evidence to indicate that the standard deviation of hole diameter exceeds 0.01 millimeters.

your answer is 23/2 - 13.5

step-by-step explanation:

7/2 evaluates to

multiply x and 4.5

multiply x and 1

the x just gets copied along.

the answer is x

x

4.5*x evaluates to 4.5x

multiply 3 and 4.5x

multiply 1 and x

the x just gets copied along.

x

4.5*x*3 evaluates to 13.5x

7/2-4.5*x*3 evaluates to

  + 8 =

the answer is

7/2-4.5*x*3+8 evaluates to

answer:   rlly

step-by-step explanation: