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3 years ago

13. Interpret the phrase ""margin of sampling error is ±1 percentage point.""

Mathematics

1 answer:

Ulleksa [173]3 years ago

4 0

Answer:

On this case a margin of error of $\pm 1 \%$ means that the true population proportion is 1% above or below the estimated proportion calculated from the sample. And this value helps in order to find the limits for a confidence interval with a confidence given.

Step-by-step explanation:

Previous concepts

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".

The margin of error is the range of values below and above the sample statistic in a confidence interval.

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

The population proportion have the following distribution

$p \sim N(p,\sqrt{\frac{p(1-p)}{n}})$

Solution to the problem

The margin of error for the proportion interval is given by this formula:

$ME=z_{\alpha/2}\sqrt{\frac{\hat p (1-\hat p)}{n}}$ (a)

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Answer:

6.1-10.3

Step-by-step explanation:

Multiply standard deviation times 3 = 2.1

Subtract and add to mean.

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3 years ago

AC = 143, BC = 8x + 4, and AB = 7x + 4, Find BC.

dedylja [7]

Answer:

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4 years ago

Mr. Good Wrench advertises that a customer will have to wait no more than 30 minutes for an oil change. A sample of 26 oil chang

Andru [333]

Answer:

The 90% confidence interval for the population standard deviation waiting time for an oil change is (3.9, 6.3).

Step-by-step explanation:

The (1 - α)% confidence interval for the population standard deviation is:

$CI=\sqrt{\frac{(n-1)s^{2}}{\chi^{2}_{\alpha/2, (n-1)}}}\leq \sigma\leq \sqrt{\frac{(n-1)s^{2}}{\chi^{2}_{1-\alpha/2, (n-1)}}}$

The information provided is:

n = 26

s = 4.8 minutes

Confidence level = 90%

Compute the critical values of Chi-square as follows:

$\chi^{2}_{\alpha/2, (n-1)}=\chi^{2}_{0.10/2, (26-1)}=\chi^{2}_{0.05, 25}=37.652$

$\chi^{2}_{1-\alpha/2, (n-1)}=\chi^{2}_{1-0.10/2, (26-1)}=\chi^{2}_{0.95, 25}=14.611$

*Use a Chi-square table.

Compute the 90% confidence interval for the population standard deviation waiting time for an oil change as follows:

$CI=\sqrt{\frac{(n-1)s^{2}}{\chi^{2}_{\alpha/2, (n-1)}}}\leq \sigma\leq \sqrt{\frac{(n-1)s^{2}}{\chi^{2}_{1-\alpha/2, (n-1)}}}$

$=\sqrt{\frac{(26-1)\times 4.8^{2}}{37.652}}\leq \sigma\leq \sqrt{\frac{(26-1)\times 4.8^{2}}{14.611}}\\\\=3.9113\leq \sigma\leq 6.2787\\\\\approx 3.9 \leq \sigma\leq6.3$

Thus, the 90% confidence interval for the population standard deviation waiting time for an oil change is (3.9, 6.3).

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3 years ago

-9. - 5 < 17 OR 13x + 25 < -1

STatiana [176]

Step-by-step explanation:

First step you take 25 on the other side and when it goes on other side it change the sign to negative 13x< -1-25
second step you keep the sign of the bigger number then you add. The result will be 13x< -26
third step you divide both side by 13x÷-26
finally the answer will be -2

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4 years ago

Question 4 of 25

Ronch [10]

Answer:

The answer is B

Step-by-step explanation:

As polynomial means "many".

So, the power of expression must be greater than 1 (e.g. x², x⁵, x⁷)

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3 years ago