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Advocard [28]
3 years ago
11

A political interest group wants to determine what fraction p ∈ (0, 1) of the population intends to vote for candidate A in the

next election. 1000 randomly chosen individuals are polled. 457 of these indicate that they intend to vote for candidate A. Find the 95% confidence interval for the true fraction p.
Mathematics
1 answer:
mash [69]3 years ago
8 0

Answer:

95% confidence interval for the true fraction p is (0.426, 0.488)

Step-by-step explanation:

Confidence Interval can be calculated using p±ME where

  • p is the sample proportion (\frac{457}{1000} =0.457
  • ME is the margin of error from the mean

and margin of error (ME) around the mean can be found using the formula

ME=\frac{z*\sqrt{p*(1-p)}}{\sqrt{N} } where

  • z is the corresponding statistic in 95% confidence level (1.96)
  • p is the sample proportion (0.457)
  • N is the sample size (1000)

then ME=\frac{1.96*\sqrt{0.457*0.543}}{\sqrt{1000} } ≈ 0.031

Then 95% confidence interval would be 0.457±0.031 or (0.426, 0.488)

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Circumference = Diameter * \pi

So 9\pi is the answer
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a florist makes 24 bouquets she uses 16 flowers for each bouquet altogether how many flowers does she use?
NNADVOKAT [17]

Answer:

384

Step-by-step explanation:

24 bouquets * 16 flowers = 384 flowers


Hope this helps! Please mark brainliest! Have a great day!



Thank you for using Brainly!

7 0
3 years ago
Solve for b.
frutty [35]
E. None of the above. -16b = 3, divide both sides by -16, b = -3/16
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3 years ago
A random sample of 747 obituaries published recently in Salt Lake City newspapers revealed that 344 (or 46%) of the decedents di
cupoosta [38]

Answer:

The probability value is almost equal to 0. Implying that the proportion of people dying  in that particular interval if deaths occurred randomly throughout the year is unusual.

Step-by-step explanation:

The random variable <em>X</em> can be defined as the number of decedents who died in the three-month period following their birthdays.

A random sample of 747 obituaries published recently in Salt Lake City newspapers revealed that 344 (or 46%) of the decedents died in the three-month period following their birthdays (123).

The probability (p) of anyone dying in any quarter if people die randomly during the year is simply 0.25.

The random variable <em>X</em> follows a Binomial distribution with parameters n = 747 and p = 0.25.

But the sample selected is too large and the probability of success is small.

So a Normal approximation to binomial can be applied to approximate the distribution of <em>p</em> if the following conditions are satisfied:

  1. np ≥ 10
  2. n(1 - p) ≥ 10

Check the conditions as follows:

 np=747\times 0.25=186.75>10\\n(1-p)=747\times (1-0.46)=560.25>10

Thus, a Normal approximation to binomial can be applied.

So,  p\sim N(\hat p,\ \frac{\hat p(1-\hat p)}{n}).

Compute the probability that 46% or more would die in that particular interval if deaths occurred randomly throughout the year as follows:

P (p\geq 0.46)=P(\frac{p-\hat p}{\sqrt{\frac{\hat p(1-\hat p)}{n}}}>\frac{0.46-0.25}{\sqrt{\frac{0.25(1-0.25)}{747}}})

                   =P(Z>13.25)\\=1-P(Z

 *Use a <em>z</em> table for the probability.

The probability value is almost equal to 0. This probability is very low indicating that the proportion of people dying  in that particular interval if deaths occurred randomly throughout the year is unusual.

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It is 15 because I think it makes sense
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