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3 years ago

5

The boiling point of water is 212°F. The boiling point of helium is –452°F. How would you compare the temperatures using an ineq

uality?

2 answers:

KatRina [158]3 years ago

6 0

-452°F < 212°F
A negative number is always less than a positive.

Ulleksa [173]3 years ago

4 0

Answer:

-452°F < 212°F

Step-by-step explanation:

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Answer:

a) The 95% CI for the true average porosity is (4.51, 5.19).

b) The 98% CI for true average porosity is (4.11, 5.01)

c) A sample size of 15 is needed.

d) A sample size of 101 is needed.

Step-by-step explanation:

a. Compute a 95% CI for the true average porosity of a certain seam if the average porosity for 20 specimens from the seam was 4.85.

We have that to find our $\alpha$ level, that is the subtraction of 1 by the confidence interval divided by 2. So:

$\alpha = \frac{1-0.95}{2} = 0.025$

Now, we have to find z in the Ztable as such z has a pvalue of $1-\alpha$ .

So it is z with a pvalue of $1-0.025 = 0.975$ , so $z = 1.96$

Now, find the margin of error M as such

$M = z*\frac{\sigma}{\sqrt{n}}$

In which $\sigma$ is the standard deviation of the population and n is the size of the sample.

$M = 1.96*\frac{0.78}{\sqrt{20}} = 0.34$

The lower end of the interval is the sample mean subtracted by M. So it is 4.85 - 0.34 = 4.51

The upper end of the interval is the sample mean added to M. So it is 4.35 + 0.34 = 5.19

The 95% CI for the true average porosity is (4.51, 5.19).

b. Compute a 98% CI for true average porosity of another seam based on 16 specimens with a sample average of 4.56.

Following the same logic as a.

98% C.I., so $z = 2.327$

$M = 2.327*\frac{0.78}{\sqrt{16}} = 0.45$

4.56 - 0.45 = 4.11

4.56 + 0.45 = 5.01

The 98% CI for true average porosity is (4.11, 5.01)

c. How large a sample size is necessary if the width of the 95% interval is to be 0.40?

A sample size of n is needed.

n is found when M = 0.4.

95% C.I., so Z = 1.96.

$M = z*\frac{\sigma}{\sqrt{n}}$

$0.4 = 1.96*\frac{0.78}{\sqrt{n}}$

$0.4\sqrt{n} = 1.96*0.78$

$\sqrt{n} = \frac{1.96*0.78}{0.4}$

$(\sqrt{n})^{2} = (\frac{1.96*0.78}{0.4})^{2}$

$n = 14.6$

Rounding up

A sample size of 15 is needed.

d. What sample size is necessary to estimate the true average porosity to within 0.2 with 99% confidence?

99% C.I., so z = 2.575

n when M = 0.2.

$M = z*\frac{\sigma}{\sqrt{n}}$

$0.2 = 2.575*\frac{0.78}{\sqrt{n}}$

$0.2\sqrt{n} = 2.575*0.78$

$\sqrt{n} = \frac{2.575*0.78}{0.2}$

$(\sqrt{n})^{2} = (\frac{2.575*0.78}{0.2})^{2}$

$n = 100.85$

Rounding up

A sample size of 101 is needed.

8 0

4 years ago