Fizzy Waters has been involved in multiple lawsuits regarding the alkaline substance concentration in their product and getting in trouble with clients about the numerous health claims they have been making. The last thing they want to deal with is not delivering on the advertised amount of content. They want to ensure that the filling process of their 160 oz (1.25 gal) bottles of alkaline water is not highly variable. The quality control engineer at Fizzy Waters suggests repairing the filling machine if the variance of the process is more than 25 oz. 8 random samples were collected and their content in ounces (oz) is given in the table below. 158.2 162.8 161.5 161.2 166.5 160.1
158.4 175.6 159.9 168.8 161.9 163.7
Perform a hypothesis test with 90% reliability to determine whether the filling machine needs to be repaired. Based on the test, explain why the filling machine needs or does not need to be repaired.

Step-by-step explanation:

For the sample, n = 12

Mean, x = (158.2 + 162.8 + 161.5 + 161.2 + 166.5 + 160.1 + 158.4 + 175.6 + 159.9 + 168.8 + 161.9 + 163.7)/12 = 163.22

Variance, s² = (summation(x - mean)²/n

Summation(x - mean)² = (158.2 - 163.22)^2 + (162.8 - 163.22)^2 + (161.5 - 163.22)^2 + (161.2 - 163.22)^2+ (166.5 - 163.22)^2 + (160.1 - 163.22)^2 + (158.4 - 163.22)^2 + (175.6 - 163.22)^2 + (159.9 - 163.22)^2 + (168.8 - 163.22)^2 + (161.9 - 163.22)^2 + (163.7 - 163.22)^2 = 273.5368

Variance, s² = 273.5368/12 = 22.79

This is a test for a single variance. We would set up the test hypothesis.

For the null hypothesis,

H0: σ² ≥ 25

For the alternative hypothesis,

H1: σ² < 25

The formula for determining the test statistic,x² is

x² = (n - 1)s²/σ²

Where n - 1 is the degree of freedom, df.

df = 12 - 1 = 11

x² = (11 × 22.79)/25 = 10.0276

For a test of 90% reliability, Confidence level = 0.9

Cl = 1 - alpha(level of significance)

alpha = 1 - 0.9 = 0.1

The critical value from the chi-square distribution table is 17.28. Since 17.28 > 10.0276, we would reject the null hypothesis. Therefore, the filling machine does not need repair because the variance of the process is not more than 25 oz.

what?

step-by-step explanation:

what do you need with. i love quadrants. can you tell me ?

part a

answer: a reasonable domain is where n is a real number. so n can be between 0 and 10. both endpoints are included.

work shown:

n is the number of days after the experiment starts. the smallest n can be is n = 0 which means that 0 days have gone by, and we're at the start. to find out how large n should be, then replace f(n) with 16.13 and solve for n. use logarithms to isolate the exponent.

f(n) = 12*(1.03)^n

16.13 = 12*(1.03)^n

16.13/12 = (1.03)^n

1.34416666666667 = (1.03)^n

(1.03)^n = 1.34416666666667

log[ (1.03)^n ] = log[ 1.34416666666667 ]

n*log[ 1.03 ] = log[ 1.34416666666667 ]

n = log[ 1.34416666666667 ]/log[ 1.03 ]

n = 10.0062999823929

this rounds to n = 10 which is fairly close but not 100% perfect. so this is the largest n can be.

note: if you plug n = 10 into f(n), you'll get roughly 16.127 which rounds to 16.13 (this comes up again in part c)

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part b

answer: the y intercept is 12. it represents the starting height of the plant in cm.

work shown:

plug n = 0 into the f(n) function. simplify

f(n) = 12*(1.03)^n

f(0) = 12*(1.03)^0

f(0) = 12*(1)

f(0) = 12

on day n = 0, aka the starting point, the height f(n) is 12 cm

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part c

answer: the average rate of change is approximately 0.43061036458991 (round however you need to). this represents the average growth rate from day n = 3 to day n = 10. so the plant grew roughly 0.43 cm per day during this timespan, assuming you round to 2 decimal places.

work shown:

compute f(3)

f(n) = 12*(1.03)^n

f(3) = 12*(1.03)^3

f(3) = 13.112724 < we'll use this later

compute f(10)

f(n) = 12*(1.03)^n

f(10) = 12*(1.03)^10

f(10) = 16.1269965521294 < we'll use this later

now use the formula below with a = 3 and b = 10

aroc = average rate of change

aroc = [ f(b) - f(a) ]/[ b - a ]

aroc = [ f(10) - f(3) ]/[ 10 - 3 ]

aroc = (16.1269965521294 - 13.112724)/(10 - 3)

aroc = 3.0142725521294/7

aroc = 0.43061036458991

round this however you need to

note: the plant grew approximately 3.01 cm over 7 days, so roughly 0.43 cm per day is the average growth rate (if you were to round to 2 decimal places).