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slega [8]
4 years ago
8

Wang Xii Yang has 109109109 dollars in her savings account. She has -21 dollars in her checking account. Write an inequality tha

t correctly compares the account values.
Mathematics
1 answer:
victus00 [196]4 years ago
7 0

Answer:

The answer is 109>88

Step-by-step explanation:

From the question above, we can conclude that; in her savings account, Wang Xii has $109, and in her checking account, she has -$21 less than on her saving.

Therefore, we are to find:

109 - 21=88

The answer is 109>88.

This simply means that she has more money in her savings account.

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Rudik [331]

Answer:

a) 0.93 - 1.96\sqrt{\frac{0.93(1-0.93)}{500}}=0.908

0.93 + 1.96\sqrt{\frac{0.93(1-0.93)}{500}}=0.952

The 95% confidence interval would be given by (0.908;0.0.952)

b) 0.21 - 2.58\sqrt{\frac{0.21(1-0.21)}{500}}=0.163

0.21 + 2.58\sqrt{\frac{0.21(1-0.21)}{500}}=0.257

The 99% confidence interval would be given by (0.163;0.0.257)

c) The margin of error for part a is:

ME= 1.96\sqrt{\frac{0.93(1-0.93)}{500}}=0.0224

And for part b is:

ME=2.58\sqrt{\frac{0.21(1-0.21)}{500}}=0.0470

So then the margin of error is larger for part b.

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 a

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:

z_{\alpha/2}=-1.96, z_{1-\alpha/2}=1.96

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

\hat p \pm z_{\alpha/2}\sqrt{\frac{\hat p (1-\hat p)}{n}}

If we replace the values obtained we got:

0.93 - 1.96\sqrt{\frac{0.93(1-0.93)}{500}}=0.908

0.93 + 1.96\sqrt{\frac{0.93(1-0.93)}{500}}=0.952

The 95% confidence interval would be given by (0.908;0.0.952)

Part b

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 99% of confidence, our significance level would be given by \alpha=1-0.99=0.01 and \alpha/2 =0.005. And the critical value would be given by:

z_{\alpha/2}=-2.58, z_{1-\alpha/2}=2.58

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

\hat p \pm z_{\alpha/2}\sqrt{\frac{\hat p (1-\hat p)}{n}}

If we replace the values obtained we got:

0.21 - 2.58\sqrt{\frac{0.21(1-0.21)}{500}}=0.163

0.21 + 2.58\sqrt{\frac{0.21(1-0.21)}{500}}=0.257

The 99% confidence interval would be given by (0.163;0.0.257)

Part c

The margin of error for part a is:

ME= 1.96\sqrt{\frac{0.93(1-0.93)}{500}}=0.0224

And for part b is:

ME=2.58\sqrt{\frac{0.21(1-0.21)}{500}}=0.0470

So then the margin of error is larger for part b.

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