Question

1. A lumber company is interested in assessing the precision of a new machine. The cut width is set to 1.5" and a random sample of size n = 15 is collected to assess the cutting accuracy of the machine. It is known that the standard deviation is σ = 0.05 inches.  

What is the Null Hypothesis?

2.

A lumber company is interested in assessing the precision of a new machine. The cut width is set to 1.5" and a random sample of size n = 15 is collected to assess the cutting accuracy of the machine. It is known that the standard deviation is σ = 0.05 inches.  

Using a rejection criteria of x ¯ < 1.475 , what is the Type-I error rate?

Round your answer to the nearest thousandth (3 decimal places).

3.

A lumber company is interested in assessing the precision of a new machine. The cut width is set to 1.5" and a random sample of size n = 15 is collected to assess the cutting accuracy of the machine. It is known that the standard deviation is σ = 0.05 inches.  

Using a rejection criteria of x ¯ < 1.475 , what is the Type-II error rate if the true mean cutting width is μ = 1.46 ?

Round your answer to the nearest thousandth (3 decimal places).

Answer 1

0.05 Given that, n=15, o= Null Hypothesis :- Ho: M = 1.5 9.2. Using rejection criteria of x < 1.475, the type I error rate true P [ Reject Hol Ho is [x<1.445 / 4 =15] =P PE - lun 1.475-105 0.05 1515 where ZNN(0,1) P(Z < -1.9365] = NORMSDIST (-109365) 0.026 - { using mus-Exal} Type I enror rate 0.026 9.3. Using error rate the true : Type II esmor rate rejection criteria of n < 1.475, the type II if mean cutting width is l=1046. P[ Accept Ho I Hi is true EP [> 1.445 | 451:46 PLE 1.475-1046 0.05/115 Ep[2 > 11619] where ZmN10,r) - I-P[Z 21.1619] El- NORMS OIST (1:1619) O. 123 { Using MS-Excep} = P you こ ... Type II esmor rate 10.123