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3 years ago

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A rectangular tank is 100 cm long, 30 cm wide and 12 cm deep. What is the estimated capacity of the tank half full? Hint: 1000 c

ubic centimeter = 1Litre *

1 answer:

Licemer1 [7]3 years ago

3 0

To figure this out, you need to multiply all the dimensions together. 100x30=. 3,000cm and 3,000x12= 36,000 cubic centimeters. So now you can divide the 36,000 by 1000 to get the amount of liters the entire tank can hold. 36,000/1000= 36 liters. And because you want to know how much half of the tank is, you divide 36 by 2 and get 18 liters

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Suppose we are interested in analyzing the weights of NFL players. We know that on average, NFL players weigh 247 pounds with a

tigry1 [53]

Answer:

$SE= \frac{\sigma}{\sqrt{n}} = \frac{47}{\sqrt{30}}= 8.58$

$ME= 1.64 *\frac{\sigma}{\sqrt{n}} = 1.64*\frac{47}{\sqrt{30}}= 14.073$

$\bar X - ME = 237- 14.073 = 222.927$

$\bar X + ME = 237+ 14.073 = 251.073$

$n=(\frac{1.640(47)}{5})^2 =237.65 \approx 238$

So the answer for this case would be n=238 rounded up to the nearest integer

Null hypothesis: $\mu \geq 247$

Alternative hypothesis: $\mu$

$z=\frac{237-247}{\frac{47}{\sqrt{30}}}=-1.165$

$p_v =P(z$

If we compare the p value and the significance level given $\alpha=0.1$ we see that $p_v>\alpha$ so we can conclude that we have enough evidence to FAIL to reject the null hypothesis, and we can't conclude that the true mean is lower than 247 at 10% of significance.

Step-by-step explanation:

For this case we have the following data given:

$\bar X =237$ represent the sample mean

$\sigma = 47$ represent the population deviation

$n =30$ represent the sample size selected

$\mu_0 = 247$ represent the value that we want to test.

The standard error for this case is given by:

$SE= \frac{\sigma}{\sqrt{n}} = \frac{47}{\sqrt{30}}= 8.58$

For the 90% confidence the value of the significance is given by $\alpha=1-0.9 = 0.1$ and $\alpha/2 = 0.05$ so we can find in the normal standard distribution a quantile that accumulates 0.05 of the area on each tail and we got:

$z_{\alpha/2}= 1.64$

And the margin of error would be:

$ME= 1.64 *\frac{\sigma}{\sqrt{n}} = 1.64*\frac{47}{\sqrt{30}}= 14.073$

The confidence interval for this case would be given by:

$\bar X - ME = 237- 14.073 = 222.927$

$\bar X + ME = 237+ 14.073 = 251.073$

The margin of error is given by this formula:

$ME=z_{\alpha/2}\frac{\sigma}{\sqrt{n}}$ (a)

And on this case we have that ME =5 and we are interested in order to find the value of n, if we solve n from equation (a) we got:

$n=(\frac{z_{\alpha/2} \sigma}{ME})^2$ (b)

Replacing into formula (b) we got:

$n=(\frac{1.640(47)}{5})^2 =237.65 \approx 238$

So the answer for this case would be n=238 rounded up to the nearest integer

State the null and alternative hypotheses.

We need to conduct a hypothesis in order to check if the true mean is lower than 247 pounds, the system of hypothesis would be:

Null hypothesis: $\mu \geq 247$

Alternative hypothesis: $\mu$

Since we know the population deviation, is better apply a z test to compare the actual mean to the reference value, and the statistic is given by:

$z=\frac{\bar X-\mu_o}{\frac{\sigma}{\sqrt{n}}}$ (1)

z-test: "Is used to compare group means. Is one of the most common tests and is used to determine if the mean is (higher, less or not equal) to an specified value".

Calculate the statistic

We can replace in formula (1) the info given like this:

$z=\frac{237-247}{\frac{47}{\sqrt{30}}}=-1.165$

P-value

Since is a left tailed test the p value would be:

$p_v =P(z$

Conclusion

If we compare the p value and the significance level given $\alpha=0.1$ we see that $p_v>\alpha$ so we can conclude that we have enough evidence to FAIL to reject the null hypothesis, and we can't conclude that the true mean is lower than 247 at 10% of significance.

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Step-by-step explanation:

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