Answer the following question CORRECTLY I will know if this is wrong. I WILL REPORT ANY INCORRECT ANSWERS!

Mathematics

2 answers:

lana66690 [7]3 years ago

8 0

Answer: A)toy cars

Why I think this: Think about what will be placed in the toy box. Mr. Winters can best use the toy box for toy cars.

gogolik [260]3 years ago

6 0

A.) Toy cars

That is because since Mr. Winters is going to buy a toy box that can hold as many toys as the one be replaced, he can use toy cars to measure or determine the size of the toy box.

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1. A 2002 poll reported that 61% of people worried that they would be exposed to SARS.

Mama L [17]

Answer:

Part 1

a) $ME=1.96 \sqrt{\frac{0.61 (1-0.61)}{100}}=0.0956$

b) $ME=1.96 \sqrt{\frac{0.61 (1-0.61)}{363}}=0.0502$

c) $ME=1.96 \sqrt{\frac{0.61 (1-0.61)}{1551}}=0.0243$

Part 2

We can see that if we increase the sample size the margin of error decrease and that makes sense since n is on the denominator in the formula for the margin of error and if we increase the denominator the result needs to decrease.

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

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

Part 1

For this case in order to find the margin of error we need to assume a confidence level fixed, let's assume 95% for example. The Margin of error is given by this formula:

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

In order to find the critical value we need to take in count that we are finding the interval for a proportion, so on this case we need to use the z distribution. Since our interval is at 95% of confidence, our significance level would be given by $\alpha=1-0.95=0.05$ and $\alpha/2 =0.025$ . And the critical value would be given by: