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Pepsi [2]
3 years ago
6

How many tissues should the kimberly clark corporation package of kleenex contain? researchers determined that 60 tissues is the

mean number of tissues used during a cold. suppose a random sample of 100 kleenex users yielded a mean of 52 tissues used during a cold with a standard deviation of 22. suppose the alternative you wanted to test was h1: µ < 60. if the critical value in this situation at an α = .05 is -1.6604, when should we reject the null?
Mathematics
1 answer:
Agata [3.3K]3 years ago
5 0

Answer:

Step-by-step explanation:

For the null hypothesis,

µ = 60

For the alternative hypothesis,

h1: µ < 60

This is a left tailed test

Since the population standard deviation is not given, the distribution is a student's t.

Since n = 100,

Degrees of freedom, df = n - 1 = 100 - 1 = 99

t = (x - µ)/(s/√n)

Where

x = sample mean = 52

µ = population mean = 60

s = samples standard deviation = 22

t = (52 - 60)/(22/√100) = - 3.64

We would determine the p value using the t test calculator. It becomes

p = 0.00023

We would reject the null hypothesis if α = 0.05 > 0.00023

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9.2x10^-8

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How many students were surveyed?<br>​
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20

Step-by-step explanation:

count the dots

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Write down the explicit solution for each of the following: a) x’=t–sin(t); x(0)=1
Kay [80]

Answer:

a) x=(t^2)/2+cos(t), b) x=2+3e^(-2t), c) x=(1/2)sin(2t)

Step-by-step explanation:

Let's solve by separating variables:

x'=\frac{dx}{dt}

a)  x’=t–sin(t),  x(0)=1

dx=(t-sint)dt

Apply integral both sides:

\int {} \, dx=\int {(t-sint)} \, dt\\\\x=\frac{t^2}{2}+cost +k

where k is a constant due to integration. With x(0)=1, substitute:

1=0+cos0+k\\\\1=1+k\\k=0

Finally:

x=\frac{t^2}{2} +cos(t)

b) x’+2x=4; x(0)=5

dx=(4-2x)dt\\\\\frac{dx}{4-2x}=dt \\\\\int {\frac{dx}{4-2x}}= \int {dt}\\

Completing the integral:

-\frac{1}{2} \int{\frac{(-2)dx}{4-2x}}= \int {dt}

Solving the operator:

-\frac{1}{2}ln(4-2x)=t+k

Using algebra, it becomes explicit:

x=2+ke^{-2t}

With x(0)=5, substitute:

5=2+ke^{-2(0)}=2+k(1)\\\\k=3

Finally:

x=2+3e^{-2t}

c) x’’+4x=0; x(0)=0; x’(0)=1

Let x=e^{mt} be the solution for the equation, then:

x'=me^{mt}\\x''=m^{2}e^{mt}

Substituting these equations in <em>c)</em>

m^{2}e^{mt}+4(e^{mt})=0\\\\m^{2}+4=0\\\\m^{2}=-4\\\\m=2i

This becomes the solution <em>m=α±βi</em> where <em>α=0</em> and <em>β=2</em>

x=e^{\alpha t}[Asin\beta t+Bcos\beta t]\\\\x=e^{0}[Asin((2)t)+Bcos((2)t)]\\\\x=Asin((2)t)+Bcos((2)t)

Where <em>A</em> and <em>B</em> are constants. With x(0)=0; x’(0)=1:

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7 0
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Answer:

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