When testing gas pumps for​ accuracy, fuel-quality enforcement specialists tested pumps and found that of them were not pumping accurately​ (within 3.3 oz when 5 gal is​ pumped), and pumps were accurate. Use a 0.01 significance level to test the claim of an industry representative that less than​ 20% of the pumps are inaccurate. Use the​ P-value method and use the normal distribution as an approximation to the binomial distribution. Identify the null hypothesis and alternative hypothesis. A. ​: p0.2 ​: p0.2 B. ​: p0.2 ​: p0.2 C. ​: p0.2 ​: p0.2 D. ​: p0.2 ​: p0.2 E. ​: p0.2 ​: p0.2 F. ​: p0.2 ​: p0.2 The test statistic is z nothing. ​(Round to four decimal places as​ needed.) The​ P-value is nothing. ​(Round to four decimal places as​ needed.) Because the​ P-value is ▼ greater than less than the significance​ level, ▼ fail to reject reject the null hypothesis. There is ▼ insufficient sufficient evidence support the claim that less than​ 20% of the pumps are inaccurate.

Answers

Answer 1

Complete question :

When testing gas pumps for​ accuracy, fuel-quality enforcement specialists tested pumps and found that 1294 of them were not pumping accurately​ (within 3.3 oz when 5 gal is​ pumped), and 5705 pumps were accurate. Use a 0.01 significance level to test the claim of an industry representative that less than​ 20% of the pumps are inaccurate. Use the​ P-value method and use the normal distribution as an approximation to the binomial distribution.

Answer:

H0​: p=0.2

H1​: p<0.2 ;

- 3.16 ;

0.0008 ;

p value is < α ; We reject the Null.

Therefore we conclude that there is sufficient evidence to support the claim that less than 20% of the pumps are inaccurate

Step-by-step explanation:

The null hypothesis ; H0 : p = 20%

Alternative ; H1 : p < 20%

Sample size, n = 1294 + 5705 = 6999

p = inaccurate / total

p = 1294 / 6999 = 0.1849

The test statistic (Z) :

(P - P0) ÷ sqrt((P0(1 - P0)) /n)

P0 = 0.2

(0.1849 - 0.2) ÷ sqrt((0.2 * 0.8)/6999))

−0.0151 ÷ 0.0047812

= - 3.1582029

= - 3.16

Pvalue : using the Pvalue from Zstatistic calculator at α = 0.01

P value = 0.000789

P value = 0.0008 (4 decimal places).

Since p value is < α ; We reject the Null.

Therefore we conclude that there is sufficient evidence to support the claim that less than 20% of the pumps are inaccurate


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Answers

Answer:

[tex]\displaystyle s = \frac{5t^4}{4} + \frac{9}{t} - \frac{9}{4}[/tex]

General Formulas and Concepts:

Pre-Algebra

Order of Operations: BPEMDAS

BracketsParenthesisExponentsMultiplicationDivisionAdditionSubtractionLeft to Right  

Equality Properties

Multiplication Property of EqualityDivision Property of EqualityAddition Property of EqualitySubtraction Property of Equality

Algebra I

Exponential Rule [Rewrite]:                                                                           [tex]\displaystyle b^{-m} = \frac{1}{b^m}[/tex]

Calculus

Derivatives

Derivative Notation

Solving Differentials - Integrals

Integration Constant C

Integration Rule [Reverse Power Rule]:                                                               [tex]\displaystyle \int {x^n} \, dx = \frac{x^{n + 1}}{n + 1} + C[/tex]

Integration Property [Multiplied Constant]:                                                         [tex]\displaystyle \int {cf(x)} \, dx = c \int {f(x)} \, dx[/tex]

Integration Property [Addition/Subtraction]:                                                       [tex]\displaystyle \int {[f(x) \pm g(x)]} \, dx = \int {f(x)} \, dx \pm \int {g(x)} \, dx[/tex]

Step-by-step explanation:

*Note:

Ignore the Integration Constant C on the left hand side of the differential equation when integrating.

Step 1: Define

[tex]\displaystyle \frac{ds}{dt} = 5t^3 + \frac{9}{t^2}[/tex]

t = 1

s = 8

Step 2: Integrate

[Derivative] Rewrite [Leibniz's Notation]:                                                     [tex]\displaystyle ds = (5t^3 + \frac{9}{t^2})dt[/tex][Equality Property] Integrate both sides:                                                     [tex]\displaystyle \int {} \, ds = \int {(5t^3 + \frac{9}{t^2})} \, dt[/tex][Left Integral] Reverse Power Rule:                                                             [tex]\displaystyle s = \int {(5t^3 + \frac{9}{t^2})} \, dt[/tex][Right Integral] Rewrite [Integration Property - Addition]:                           [tex]\displaystyle s = \int {5t^3} \, dt + \int {\frac{9}{t^2}} \, dt[/tex][Right Integrals] Rewrite [Integration Property - Multiplied Constant]:     [tex]\displaystyle s = 5\int {t^3} \, dt + 9\int {\frac{1}{t^2}} \, dt[/tex][Right Integrals] Rewrite [Exponential Rule - Rewrite]:                               [tex]\displaystyle s = 5\int {t^3} \, dt + 9\int {t^{-2}} \, dt[/tex][Right Integrals] Reverse Power Rule:                                                         [tex]\displaystyle s = 5(\frac{t^4}{4}) + 9(\frac{t^{-1}}{-1}) + C[/tex][Right Integrals] Rewrite [Exponential Rule - Rewrite]:                               [tex]\displaystyle s = 5(\frac{t^4}{4}) + 9(\frac{1}{t}) + C[/tex]Multiply:                                                                                                         [tex]\displaystyle s = \frac{5t^4}{4} + \frac{9}{t} + C[/tex]

Step 3: Solve

Substitute in variables:                                                                                 [tex]\displaystyle 8 = \frac{5(1)^4}{4} + \frac{9}{1} + C[/tex]Evaluate exponents:                                                                                     [tex]\displaystyle 8 = \frac{5}{4} + \frac{9}{1} + C[/tex]Divide:                                                                                                           [tex]\displaystyle 8 = \frac{5}{4} + 9 + C[/tex]Add:                                                                                                               [tex]\displaystyle 8 = \frac{41}{4} + C[/tex][Subtraction Property of Equality] Isolate C:                                               [tex]\displaystyle \frac{-9}{4} = C[/tex]Rewrite:                                                                                                          [tex]\displaystyle C = \frac{-9}{4}[/tex]

Particular Solution: [tex]\displaystyle s = \frac{5t^4}{4} + \frac{9}{t} - \frac{9}{4}[/tex]

Topic: AP Calculus AB/BC (Calculus I/II)

Unit: Differentials Equations and Slope Fields

Book: College Calculus 10e

Can anybody help me

Answers

Since there is no value of x that will ever make this a true statement, the solution to the equation above is “no solution”  The solution x = 0 means that the value 0 satisfies the equation, so there is a solution. “No solution” means that there is no value, not even 0, which would satisfy the equation. and Basically just means for all numbers from minus infinity to infinity. We say real x to distinguish from complex numbers. 0. general ebriety. and Basically just means for all numbers from minus infinity to infinity. We say real x to distinguish from complex numbers. 0. general ebriety.

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