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andrezito [222]

3 years ago

6

Use technology to find the P-value for the hypothesis test described below. The claim is that for 12 AM body temperatures, the

mean is mu μ greater than > 98.6 98.6 degrees °F. The sample size is n equals = 9 9 and the test statistic is t equals = 2.132 2.132.

1 answer:

Viktor [21]3 years ago

4 0

Answer:

P-value = 0.032794

Step-by-step explanation:

We are given the following information in the question:

Population mean, μ = 98.6 degrees

Sample size, n = 9

Alpha, α = 0.05

Test t-statistic = 2.132

The null and the alternate hypothesis :

$H_{0}: \mu = 98.6\text{ degrees}\\H_A: \mu > 98.6\text{ degrees}$

We have to find the p-value for degree of freedom 8 and significance level 0.05

The calculated p-value is 0.032794

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Find the function y = f(t) passing through the point (0, 18) with the given first derivative.

monitta

Answer:

$\displaystyle y = \frac{t^2}{16} + 18$

General Formulas and Concepts:

<u>Pre-Algebra</u>

Order of Operations: BPEMDAS

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Parenthesis
Exponents
Multiplication
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Left to Right

Equality Properties

Multiplication Property of Equality
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Addition Property of Equality
Subtraction Property of Equality

<u>Algebra I</u>

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Function Notation
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<u>Calculus</u>

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Derivative Notation

Antiderivatives - Integrals

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

Step-by-step explanation:

<u>Step 1: Define</u>

<em>Identify</em>

Point (0, 18)

$\displaystyle \frac{dy}{dt} = \frac{1}{8} t$

<u>Step 2: Find General Solution</u>

<em>Use integration</em>

[Derivative] Rewrite: $\displaystyle dy = \frac{1}{8} t\ dt$
[Equality Property] Integrate both sides: $\displaystyle \int dy = \int {\frac{1}{8} t} \, dt$
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Substitute in point [Function]: $\displaystyle 18 = \frac{0^2}{16} + C$
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Substitute in <em>C</em> [Function]: $\displaystyle y = \frac{t^2}{16} + 18$

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

Unit: Integration

Book: College Calculus 10e

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