The quality manager at Bestiful Company is certifying a new process that must produce 95% (or better) good product before the process is considered proven. A sample of 40 containers from the process line are tested, and 93% are found to be good. Suppose the null hypothesis is formulated as: p = 95%, and the alternative hypothesis is formulated as: p < 95%. The significance level is set at α = 0.01. Calculate the p-value of the test. Please type your answer with 4 decimal places.

The calculated value Z= -0.5822

|z| = |-0.5822|< 2.576 at 0.01 level of significance

Hence null hypothesis is accepted

The quality manager at Restful Company is certifying a new process that must produce 95% (or better) good product

Step-by-step explanation:

Explanation:-

Step:-(i)

The Population proportion is P = 95% =0.95

A sample of 40 containers from the process line are tested, and 93% are found to be good

Sample size ( n) =40

The sample proportion 'p' = 0.93

Level of significance (α )= 0.01.

Null hypothesis:-p = 95%=0.95

Alternative hypothesis:- p≠0.95

The test statistic

                                            (i)

Step:(ii)

substitute all values in (i)

Z = -0.5822

The calculated value Z= -0.5822

|z| = |-0.5822|< 2.576 at 0.01 level of significance

Hence null hypothesis is accepted

Conclusion:-

The quality manager at Restful Company is certifying a new process that must produce 95% (or better) good product

(f - g)(x) = 5x^3 - 3

step-by-step explanation:

rearrange these functions vertically, with like terms in separate columns:

f(x) = 5x^3 - 2

g(x) =         +1

now subtract, column by column:

(f - g)(x) = 5x^3 - 3

step-by-step :