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Andreas93 [3]
1 year ago
5

The results for all 10 grocery surveys were actually generated using a mathematical simulation, found on the Simulation tab of t

he spreadsheet. To use it, enter a population mean and population standard deviation in the two yellow cells at the top of worksheet and follow the instructions. Every time a new set of data is generated, the data changes, providing a new sample. You can copy this data and paste it into the histogram tool to analyze it. (You don't have to do that in this activity, but it's there for you to experiment with, if you like.) Any given set of sample data generated in this way would not typically have the same mean, distribution, and standard deviation as the population values, but the simulated sample data would be consistent with a randomly selected sample from such a population. Note that the simulated population mean is 1.73, and the simulated standard deviation is 0.657. Imagine that this random sample survey was simulated an infinite number of times. In that case, the population mean (1) of all the samples would be $1.73. All the sample means would be normally distributed, according to the central limit theorem. Question 1 If the standard deviation of the population is $0.657 (as it is in the simulation), what is the standard error of the mean? Round up your answer to the nearest tenth of a cent.
Mathematics
1 answer:
Debora [2.8K]1 year ago
6 0

The standard error of the mean is given by:

\frac{\sigma}{\sqrt[]{n}}

where sigma is the standard deviation and n is the sample size. In this case the sample size is 10 and the standard deviation is 0.657, then the standard error is:

\frac{0.657}{\sqrt[]{10}}=0.208

Therefore the standard error is $0.208.

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Anvisha [2.4K]
Hello!

Reading these types of graphs can be difficult, but it displays the following numbers:

27, 30, 31, 43, 45, 49, 51, 58, 62, 64

To find mean, or average, you need to add up all of the numbers and divide by how many there are (10)

This will give you 46.

I hope this helps you and good luck!
8 0
3 years ago
Read 2 more answers
A researcher reports a 98% confidence interval for the proportion of Drosophila in a population with mutation Adh-F to be [0.34,
Fantom [35]

Answer:

0.34 \leq p \leq 0.38

For this case we can conclude that with 98% of confidence the true proportion of Drosophila in a population would be between 0.34 and 0.38.

But that doesn't means that we have 98% of PROBABILITY that the true proportion would be between 0.34 and 0.38, because we are constructing a confidence interval with sample data and we can't analyze this using probability.

Then the best answer is:

2. False

Step-by-step explanation:

For this case we have a confidence interval for the proportion of Drosophila in a population with mutation Adh-F to be given by:

0.34 \leq p \leq 0.38

For this case we can conclude that with 98% of confidence the true proportion of Drosophila in a population would be between 0.34 and 0.38.

But that doesn't means that we have 98% of PROBABILITY that the true proportion would be between 0.34 and 0.38, because we are constructing a confidence interval with sample data and we can't analyze this using probability.

Then the best answer is:

2. False

3 0
3 years ago
Please can I have some help with these?
LenaWriter [7]
I hope this helps you

4 0
3 years ago
Read 2 more answers
A newspaper reported that 4040​% of people say that some coffee shops are overpriced. The source of this information was a telep
vampirchik [111]

Answer:

a. Identify the population of interest in this study.

  • all the population of the country

b. Identify the sample for the study.

  • 50 adults (the ones that were surveyed)

c. Identify the parameter of interest in the study.

  • the parameter of interest is the people that believe that coffee shops are overpriced

d. Find and interpret a 95​% confidence interval for the parameter of interest.

z score for a 95% confidence level = 1.96

margin of error (E) = z score x √{[0.4(1 - 0.4)] / 50} = 1.96 x 0.069282 = 0.13579

95% confidence level interval:

0.4 - 0.13579 = 0.26421

0.4 + 0.13579 = 0.53579

  • 95% confidence interval (0.26421 , 0.53579)

5 0
3 years ago
The average number of acres burned by all wildfires in the united states is 780 acres with standard deviation 500 acres. of cour
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