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3 years ago

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Jane babysits for $8 per hour. She earned $240 last week. This equation represents the number of hours, h, Jane babysat last wee

k.
8h=240
How many hours did Jane babysit last week?

1 answer:

Tatiana [17]3 years ago

8 0

Answer:

30 Hours

Step-by-step explanation:

8h=240

240÷8=30

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An SRS of 350 high school seniors gained an average of x¯=21 points in their second attempt at the SAT Mathematics exam. Assume

Lynna [10]

Answer:

a) $21-2.58\frac{52}{\sqrt{350}}=13.83$

$21+2.58\frac{52}{\sqrt{350}}=28.17$

So on this case the 99% confidence interval would be given by (13.83;28.17)

b) $ME=2.58\frac{52}{\sqrt{350}}=7.17$

c) $ME=2.58\frac{52}{\sqrt{100}}=13.42$

d) D. Decreasing the sample size increases the margin of error, provided the confidence level and population standard deviation remain the same.

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".

Part a

$\bar X=21$ represent the sample mean

$\mu$ population mean (variable of interest)

$\sigma=52$ represent the population standard deviation

n=350 represent the sample size

99% confidence interval

The confidence interval for the mean is given by the following formula:

$\bar X \pm z_{\alpha/2}\frac{\sigma}{\sqrt{n}}$ (1)

Since the Confidence is 0.99 or 99%, the value of $\alpha=0.01$ and $\alpha/2 =0.005$ , and we can use excel, a calculator or a table to find the critical value. The excel command would be: "=-NORM.INV(0.005,0,1)".And we see that $z_{\alpha/2}=2.58$

Now we have everything in order to replace into formula (1):

$21-2.58\frac{52}{\sqrt{350}}=13.83$

$21+2.58\frac{52}{\sqrt{350}}=28.17$

So on this case the 99% confidence interval would be given by (13.83;28.17)

Part b

The margin of error is given by:

$ME=2.58\frac{52}{\sqrt{350}}=7.17$

Part c

The margin of error is given by:

$ME=2.58\frac{52}{\sqrt{100}}=13.42$

Part d

As we can see when we reduce the sample size we increase the margin of error so the best option for this case is:

D. Decreasing the sample size increases the margin of error, provided the confidence level and population standard deviation remain the same.

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