1. (20 points) One step in the manufacture of a certain metal clamp involves the drilling of four holes. In a sample of 150 clamps, the average time needed to complete this step was 72 seconds and standard deviation was 10 seconds. (a) Find a 99.5% confidence interval for the mean time needed to complete the step. (b) What is the confidence level of the interval (71, 73)? (c) How many clamps must be sampled so that 99.5% confidence interval specifies the mean to within ±1.5 seconds? (d) Can you conclude that the mean time needed to complete the step is greater than 73 seconds? i. State hypothesis. ii. Compute P-value. iii. What is your conclusion?

(69.708 ; 74.292) ; 78% ; 350 ;

H0 : μ = 72

H0 : μ > 72

Pvalue = 0.8897

there is not enough evidence to support the claim that the mean time needed to complete the step is greater than 73.

Step-by-step explanation:

99.5% confidence level, = 2.807

Sample size, n = 150

xbar = 72

Standard deviation, s = 10

Xbar ± Margin of error

Margin of Error = Zcritical * s/sqrt(n)

Margin of Error = 2.807 * 10/sqrt(150)

Margin of Error = 2.292

Lower boundary = 72 - 2.292 = 69.708

Upper boundary = 72 + 2.292 = 74.292

(69.708 ; 74.292)

B.)

Confidence level of the interval (71, 73)

(72 - 71 ) or (73 - 72)

Margin of Error = 1

Margin of Error = Zcritical * s/sqrt(n)

1 = Zcritical * 10/sqrt(150)

1 = Zcritical * 0.8164965

Zcritical = 1 / 0.8164965

Zcritical = 1.2247

α = 0.22 = 1 - 0.22 = 0.78 = 78%

C.)

Margin of Error = ±1.5

1.5 = Zcritical * s/sqrt(n)

99.5% confidence level, = 2.807

1.5 = 2.807 * 10/sqrt(n)

1.5 = 28.07 / sqrt(n)

Square both sides

1.5² = 28.08² / n

2.25n = 788.4864

n = 788.4864 / 2.25

n = 350.4384

n = 350 samples

D.)

H0 : μ = 72

H0 : μ > 72

Test statistic :

(xbar - μ) ÷ s/sqrt(n)

(73 - 72) ÷ 10/sqrt(150)

Test statistic = 1.2247

Pvalue :

P(Z < 1.2247) = 0.8897

Pvalue > α

0.8897 > 0.005

We fail to reject the Null ; Hence there is not enough evidence to support the claim that the mean time needed to complete the step is greater than 73.

1gm = 1000mg so 0.1gm = 1000mg

first, convert the weight to kg

160lb * 1kg/2.2lb = 72.7kg

next, the dose

72.7kg * 100mg/kg = 7270mg

option: d is correct

step-by-step explanation: