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ira [324]
3 years ago
12

In Case Study 19.1, we learned that about 56% of American adults actually voted in the presidential election of 1992, whereas ab

out 61% of a random sample claimed that they had voted. The size of the sample was not specified, but suppose it were based on 1600 American adults, a common size for such studies. (a) Into what interval of values should the sample proportion fall 68%, 95%, and almost all of the time? (Round your answers to three decimal places.) 68% to 95% to Almost all to (b) Is the observed value of 61% reasonable, based on your answer to par
Mathematics
1 answer:
Radda [10]3 years ago
7 0

Answer:

a) Confidence interval for 68% confidence level

= (0.548, 0.572)

Confidence interval for 95% confidence level

= (0.536, 0.584)

Confidence interval for 99.99% confidence level = (0.523, 0.598)

b) The sample proportion of 0.61 is unusual as falls outside all of the range of intervals where the sample mean can found for all 3 confidence levels examined.

c) Standardized score for the reported percentage using a sample size of 400 = 2.02

Since, most of the variables in a normal distribution should fall within 2 standard deviations of the mean, a sample mean that corresponds to standard deviation of 2.02 from the population mean makes it seem very plausible that the people that participated in this sample weren't telling the truth. At least, the mathematics and myself, do not believe that they were telling the truth.

Step-by-step explanation:

The mean of this sample distribution is

Mean = μₓ = np = 0.61 × 1600 = 976

But the sample mean according to the population mean should have been

Sample mean = population mean = nP

= 0.56 × 1600 = 896.

To find the interval of values where the sample proportion should fall 68%, 95%, and almost all of the time, we obtain confidence interval for those confidence levels. Because, that's basically what the definition of confidence interval is; an interval where the true value can be obtained to a certain level.of confidence.

We will be doing the calculations in sample proportions,

We will find the confidence interval for confidence level of 68%, 95% and almost all of the time (99.7%).

Basically the empirical rule of 68-95-99.7 for standard deviations 1, 2 and 3 from the mean.

Confidence interval = (Sample mean) ± (Margin of error)

Sample Mean = population mean = 0.56

Margin of Error = (critical value) × (standard deviation of the distribution of sample means)

Standard deviation of the distribution of sample means = √[p(1-p)/n] = √[(0.56×0.44)/1600] = 0.0124

Critical value for 68% confidence interval

= 0.999 (from the z-tables)

Critical value for 95% confidence interval

= 1.960 (also from the z-tables)

Critical values for the 99.7% confidence interval = 3.000 (also from the z-tables)

Confidence interval for 68% confidence level

= 0.56 ± (0.999 × 0.0124)

= 0.56 ± 0.0124

= (0.5476, 0.5724)

Confidence interval for 95% confidence level

= 0.56 ± (1.960 × 0.0124)

= 0.56 ± 0.0243

= (0.5357, 0.5843)

Confidence interval for 99.7% confidence level

= 0.56 ± (3.000 × 0.0124)

= 0.56 ± 0.0372

= (0.5228, 0.5972)

b) Based on the obtained intervals for the range of intervals that can contain the sample mean for 3 different confidence levels, the sample proportion of 0.61 is unusual as it falls outside of all the range of intervals where the sample mean can found for all 3 confidence levels examined.

c) Now suppose that the sample had been of only 400 people. Compute a standardized score to correspond to the reported percentage of 61%. Comment on whether or not you believe that people in the sample could all have been telling the truth, based on your result.

The new standard deviation of the distribution of sample means for a sample size of 400

√[p(1-p)/n] = √[(0.56×0.44)/400] = 0.0248

The standardized score for any is the value minus the mean then divided by the standard deviation.

z = (x - μ)/σ = (0.61 - 0.56)/0.0248 = 2.02

Standardized score for the reported percentage using a sample size of 400 = 2.02

Since, most of the variables in a normal distribution should fall within 2 standard deviations of the mean, a sample mean that corresponds to standard deviation of 2.02 from the population mean makes it seem very plausible that the people that participated in this sample weren't telling the truth. At least, the mathematics and myself, do not believe that they were telling the truth.

Hope this Helps!!!

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<h2><u>Part A:</u></h2>

Let's denote no of seats in first row with r1 , second row with r2.....and so on.

r1=5

Since next row will have 10 additional row each time when we move to next row,

So,

r2=5+10=15

r3=15+10=25

<u>Using the terms r1,r2 and r3 , we can find explicit formula</u>

r1=5=5+0=5+0×10=5+(1-1)×10

r2=15=5+10=5+(2-1)×10

r3=25=5+20=5+(3-1)×10

<u>So for nth row,</u>

rn=5+(n-1)×10

Since 5=r1 and 10=common difference (d)

rn=r1+(n-1)d

Since 'a' is a convention term for 1st term,

<h3><u>⇒</u><u>rn=a+(n-1)d</u></h3>

which is an explicit formula to find no of seats in any given row.

<h2><u>Part B:</u></h2>

Using above explicit formula, we can calculate no of seats in 7th row,

r7=5+(7-1)×10

r7=5+(7-1)×10 =5+6×10

r7=5+(7-1)×10 =5+6×10 =65

which is the no of seats in 7th row.

8 0
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Peter spent half the money on his gift card and coffee and he loaded another $10 on the gift card how much was on the gift card
cestrela7 [59]

Answer:

$80 was on the gift card to begin with

Step-by-step explanation:

Let the initial money be x

Peter spent (1/2)*x

Peter added 10

<em>Before adding more money on the gift card Peter had x -(1/2)x left on the gift card which is (1/2)x</em>

<em>After loading more money Peter had (1/2)x + 10</em>

(1/2)x = 40 <em>(money spent by Peter)</em>

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Answer:

Step-by-step explanation:

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Suppose that news spreads through a city of fixed size of 700000 people at a time rate proportional to the number of people who
Gnoma [55]

Answer:

y'(t) = k(700,000-y(t))  k>0 is the constant of proportionality

y(0) =0

Step-by-step explanation:

(a.) Formulate a differential equation and initial condition for y(t) = the number of people who have heard the news t days after it has happened.

If we suppose that news spreads through a city of fixed size of 700,000 people at a time rate proportional to the number of people who have not heard the news that means  

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Since no one has heard the news at first, we have

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We can then state the initial value problem as

y'(t) = k(700,000-y(t))

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