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Kipish [7]
3 years ago
10

Don tracked the price of gas for several months. In January gas cost $2.65, in February the price had risen $0.56, in March decr

eased $0.23, in April risen $1.02, and in May decreased $0.48. By how many dollars has the price of gas changed from January to May?
Mathematics
1 answer:
Elan Coil [88]3 years ago
8 0
Starting price is 2.65, increased by .56

2.65+.56=3.21

Price is 3.21, decreased by .23

3.21-.23=2.98

Price is 2.98, increased by 1.02

2.98+1.02= 4

Price is 4, decreased by .48

4-.48=3.52

Now find the difference from the starting price (January) from the end price (May)

3.52-2.65=.87

<u>The difference from gas prices is $.87</u>
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We would like to create a confidence interval.
Vlada [557]

Answer:

c.A 90% confidence level and a sample size of 300 subjects.

Step-by-step explanation:

In a sample with a number n of people surveyed with a probability of a success of \pi, and a confidence level 1-\alpha, we have the confidence interval with a margin of error of:

M = z\sqrt{\frac{\pi(1-\pi)}{n}}

In which

z is the zscore that has a pvalue of 1 - \frac{\alpha}{2}.

In this problem

The proportions are the same for all the options, so we are going to write our margins of error as functions of \sqrt{\pi(1-\pi)}

So

a.A 99% confidence level and a sample size of 50 subjects.

n = 50

99% confidence interval

So \alpha = 0.01, z is the value of Z that has a pvalue of 1 - \frac{0.01}{2} = 0.995, so Z = 2.575.

The margin of error is

M = z\sqrt{\frac{\pi(1-\pi)}{n}} = \frac{2.575}{\sqrt{50}}\sqrt{\pi(1-\pi)} = 0.3642\sqrt{\pi(1-\pi)}

b.A 90% confidence level and a sample size of 50 subjects.

n = 50

90% confidence interval

So \alpha = 0.1, z is the value of Z that has a pvalue of 1 - \frac{0.1}{2} = 0.95, so Z = 1.645.

The margin of error is

M = z\sqrt{\frac{\pi(1-\pi)}{n}} = \frac{1.645}{\sqrt{50}}\sqrt{\pi(1-\pi)} = 0.2623\sqrt{\pi(1-\pi)}

c.A 90% confidence level and a sample size of 300 subjects.

n = 300

90% confidence interval

So \alpha = 0.1, z is the value of Z that has a pvalue of 1 - \frac{0.1}{2} = 0.95, so Z = 1.645.

The margin of error is

M = z\sqrt{\frac{\pi(1-\pi)}{n}} = \frac{1.645}{\sqrt{300}}\sqrt{\pi(1-\pi)} = 0.0950\sqrt{\pi(1-\pi)}

This produces smallest margin of error.

d.A 99% confidence level and a sample size of 300 subjects.

n = 300

99% confidence interval

So \alpha = 0.01, z is the value of Z that has a pvalue of 1 - \frac{0.01}{2} = 0.995, so Z = 2.575.

The margin of error is

M = z\sqrt{\frac{\pi(1-\pi)}{n}} = \frac{2.575}{\sqrt{300}}\sqrt{\pi(1-\pi)} = 0.1487\sqrt{\pi(1-\pi)}

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Answer:−

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Step-by-step explanation:

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