Factor 7x² – 31x

Part I - To help consumers assess the risks they are taking, the Food and Drug Administration (FDA) publishes the amount of nico

IRINA_888 [86]

Answer:

(I) 99% confidence interval for the mean nicotine content of this brand of cigarette is [24.169 mg , 30.431 mg].

(II) No, since the value 28.4 does not fall in the 98% confidence interval.

Step-by-step explanation:

We are given that a new cigarette has recently been marketed.

The FDA tests on this cigarette gave a mean nicotine content of 27.3 milligrams and standard deviation of 2.8 milligrams for a sample of 9 cigarettes.

Firstly, the Pivotal quantity for 99% confidence interval for the population mean is given by;

P.Q. = $\frac{\bar X-\mu}{\frac{s}{\sqrt{n} } }$ ~ $t_n_-_1$

where, $\bar X$ = sample mean nicotine content = 27.3 milligrams

s = sample standard deviation = 2.8 milligrams

n = sample of cigarettes = 9

$\mu$ = true mean nicotine content

Here for constructing 99% confidence interval we have used One-sample t test statistics as we don't know about population standard deviation.

Part I : So, 99% confidence interval for the population mean, $\mu$ is ;

P(-3.355 < $t_8$ < 3.355) = 0.99 {As the critical value of t at 8 degree

of freedom are -3.355 & 3.355 with P = 0.5%}

P(-3.355 < $\frac{\bar X-\mu}{\frac{s}{\sqrt{n} } }$ < 3.355) = 0.99

P( $-3.355 \times {\frac{s}{\sqrt{n} } }$ < ${\bar X-\mu}$ < $3.355 \times {\frac{s}{\sqrt{n} } }$ ) = 0.99

P( $\bar X-3.355 \times {\frac{s}{\sqrt{n} } }$ < $\mu$ < $\bar X+3.355 \times {\frac{s}{\sqrt{n} } }$ ) = 0.99

99% confidence interval for $\mu$ = [ $\bar X-3.355 \times {\frac{s}{\sqrt{n} } }$ , $\bar X+3.355 \times {\frac{s}{\sqrt{n} } }$ ]

= [ $27.3-3.355 \times {\frac{2.8}{\sqrt{9} } }$ , $27.3+3.355 \times {\frac{2.8}{\sqrt{9} } }$ ]

= [27.3 $\pm$ 3.131]

= [24.169 mg , 30.431 mg]

Therefore, 99% confidence interval for the mean nicotine content of this brand of cigarette is [24.169 mg , 30.431 mg].

Part II : We are given that the FDA tests on this cigarette gave a mean nicotine content of 24.9 milligrams and standard deviation of 2.6 milligrams for a sample of n = 9 cigarettes.

The FDA claims that the mean nicotine content exceeds 28.4 milligrams for this brand of cigarette, and their stated reliability is 98%.

The Pivotal quantity for 98% confidence interval for the population mean is given by;

P.Q. = $\frac{\bar X-\mu}{\frac{s}{\sqrt{n} } }$ ~ $t_n_-_1$

where, $\bar X$ = sample mean nicotine content = 24.9 milligrams

s = sample standard deviation = 2.6 milligrams

n = sample of cigarettes = 9

$\mu$ = true mean nicotine content

Here for constructing 98% confidence interval we have used One-sample t test statistics as we don't know about population standard deviation.

So, 98% confidence interval for the population mean, $\mu$ is ;

P(-2.896 < $t_8$ < 2.896) = 0.98 {As the critical value of t at 8 degree

of freedom are -2.896 & 2.896 with P = 1%}

P(-2.896 < $\frac{\bar X-\mu}{\frac{s}{\sqrt{n} } }$ < 2.896) = 0.98

P( $-2.896 \times {\frac{s}{\sqrt{n} } }$ < ${\bar X-\mu}$ < $2.896 \times {\frac{s}{\sqrt{n} } }$ ) = 0.98

P( $\bar X-2.896 \times {\frac{s}{\sqrt{n} } }$ < $\mu$ < $\bar X+2.896 \times {\frac{s}{\sqrt{n} } }$ ) = 0.98

98% confidence interval for $\mu$ = [ $\bar X-2.896 \times {\frac{s}{\sqrt{n} } }$ , $\bar X+2.896 \times {\frac{s}{\sqrt{n} } }$ ]

= [ $24.9-2.896 \times {\frac{2.6}{\sqrt{9} } }$ , $24.9+2.896 \times {\frac{2.6}{\sqrt{9} } }$ ]

= [22.4 mg , 27.4 mg]

Therefore, 98% confidence interval for the mean nicotine content of this brand of cigarette is [22.4 mg , 27.4 mg].

No, we don't agree on the claim of FDA that the mean nicotine content exceeds 28.4 milligrams for this brand of cigarette because as we can see in the above confidence interval that the value 28.4 does not fall in the 98% confidence interval.

5 0

2 years ago

If a cylindrical tank holds 100,00 gallons of water which can be drained from thebottom of the tank in an hour, then Torricelli’

tatyana61 [14]

Answer:

First, I think the right formula is:

$V(t) = 100,000(1-\frac{t}{60} ),20\leq t\leq 60$

A) We first derive the formula given above:

$V'(t)=-\frac{100000}{60}$

V'(t) represent the drain rate of the tank volume.

B) the units is: gallons/time

C) at V'(10) = 100000/60 = 1666.66 gallons/time.

because the formula V'(t) is constant so it doesn't depend of time.

5 0

3 years ago

). If a, ß are zeroes of the quadratic polynomial p(x)=kx²+4x+4 such

natta225 [31]

Answer:

The values of k are 2/3 and -1

Step-by-step explanation:

Product of zeros = αβ= constant / coefficient of x^2 = 4/k

Sum of zeros =α+β = - coefficient of x / coefficient of x^2= -4/k

Given

Consider a= α and b= β

$(\alpha)^2 + (\beta)^2 = 24$

$(\alpha)^2 + (\beta)^2$ can be written as $(\alpha)^2 + 2(\alpha)(\beta) + (\beta)^2$ if we add $\pm 2 (\alpha)(\beta)$ in the above equation.

$(\alpha)^2 + 2(\alpha)(\beta) + (\beta)^2 -2(\alpha)(\beta)$

$(\alpha + \beta)^2 -2(\alpha)(\beta)$

Putting values of αβ and α+β

$(\frac {-4}{k})^2 -2( \frac {4}{k}) = 24\\\frac {16}{k^2} - \frac {8}{k} = 24\\Multiplying\,\, the \,\, equation\,\, with\,\, 8K^2\\ 2 - k= 3K^2\\3k^2-2+k=0\\or\\3k^2+k-2=0\\3k^2+3k-2k-2=0\\3k(k+1)-2(k+1)=0\\(3k-2)(k+1)=0\\3k-2=0 \,\,and\,\, k+1 =0\\k= 2/3 \,\,and\,\, k=-1$

The values of k are 2/3 and -1

8 0

3 years ago

Write a function rule for the table. x ƒ(x) 3 –1 4 0 5 1 6 2

MaRussiya [10]

Answer:

Step-by-step explanation:

When one variable depends on another variable according to a rule it is known as function:

For example if you want an output of 2 and the input value is 4. Then you will subtract 4 from x

We have to find x f(x).

x f(x) =?

f(x) = x-4

Now input value is: x= 3

x f(x)

3 (x-4)

3 (3-4)

3 -1

Now x= 4

x f(x)

4 (x-4)

4 (4-4)

4 0

Now x=5

x f(x)

5 (x-4)

5 (5-4)

5 1

Now x=6

x f(x)

6 (x-4)

6 (6-4)

6 2 ....

3 0

3 years ago

Which statement is true about points located in quadrant l on the coordinate plane

OverLord2011 [107]

Answer:

D

Step-by-step explanation:

Quadrant 1 is located in the upper right hand corner with x values traveling right (positive) and y values traveling up (positive)

8 0

2 years ago

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Factor 7x² – 31x – 20.