Homework Answers
Answer:
a). 95% level
2.5% from each tail
b). 90% level
5% from each tail
c). 99% level
0.5 % from each tail
d). 80% level
10% from each tail
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7. To create a confidence interval from a bootstrap distribution using percentiles, we keep the middle...
The distribution was created using 1000 bootstrap statistics. Use the distribution to estimate a? 99% confidence...
The distribution was created
using 1000 bootstrap statistics. Use the distribution to estimate
a? 99% confidence interval for the mean IQ for the population.
Round your answers to one decimal place.
The 99% confidence interval is ______ to ______.
Note - Please solve it for 99% CI, not 95%
CI.
IQ Scores A sample of 10 IQ scores was used to create the bootstrap distribution of sample means in Figure 1 # sample 1000 mean 100.104 st.dev. 4.798 80 |...
Here are the percentiles for a bootstrap distribution of average NHL Penalty minutes Percentiles for a...
Here are the percentiles for a bootstrap distribution of average NHL Penalty minutes Percentiles for a bootstrap distribution of penalty minutes 0.50% 1.00% 2.00% 2.50% 5.00% Percentile 21.8 22.8 23.9 24.4 25.8 95.00% 99.50% 97.50% 45.5 98.00% 46.1 99.00% 47.7 43.5 49.5 1. Construct a 99% Confidence interval for the true mean Penalty Minutes. 2. Construct a 98% Confidence interval for the true mean Penalty Minutes. 3. Construct a 96% Confidence interval for the true mean Penalty Minutes. 4. Construct...
confidence interval
A random sample of n = 15 heat pumps of a certain type yielded the following observations on lifetime (in years):2.0 1.1 6.0 1.7 5.3 0.4 1.0 5.315.7 0.9 4.8 0.9 12.2 5.3 0.6(a) Assume that the lifetime distribution is exponential and use an argument parallel to that of this example to obtain a 95% CI for expected (true average) lifetime.(Round your answers to two decimal places.)(b) How should the interval of part (a) be altered to achieve a confidence...
Find the indicated confidence interval. Assume the standard error comes from a bootstrap distribution that is...
Find the indicated confidence interval. Assume the standard error comes from a bootstrap distribution that is approximately normally distributed. 90% CI for mean n = 20 x = 22.9 s = 5.6 SE = 1.25 CI is ____ to ____ 95% CI n = 400 p = .4 SE = .02
Find the indicated confidence interval. Assume the standard error comes from a bootstrap distribution that is...
Find the indicated confidence interval. Assume the standard error comes from a bootstrap distribution that is approximately normally distributed. A 90% confidence interval for a mean if the sample has n = 100 with à = 22.4 and s = 5.7 , and the standard error is SE = 0.57 Round your answers to three decimal places. The 90% confidence interval is
Statistics 200: Lab Activity for Section 3.3 Constructing Bootstrap Confidence Intervals - Learning objectives: • Describe...
Statistics 200: Lab Activity for Section 3.3 Constructing Bootstrap Confidence Intervals - Learning objectives: • Describe how to select a bootstrap sample to compute a bootstrap statistic • Recognize that a bootstrap distribution tends to be centered at the value of the original statistic • Use technology to create a bootstrap distribution • Estimate the standard error of a statistic from a bootstrap distribution • Construct a 95% confidence interval for a parameter based on a sample statistic and the...
3) Find the indicated confidence interval. Assume the standard error comes from a bootstrap distribution that...
3) Find the indicated confidence interval. Assume the standard error comes from a bootstrap distribution that is approximately normally distributed. A 95% confidence interval for a difference in means μ1-μ2 if the samples have n1=80 with x¯1=269 and s1=48and n2=120 with x¯2=229 and s2=49, and the standard error is SE=6.99. Round your answers to three decimal places. The 95% confidence interval is ______________ to ____________ ______________________________________________________________________ 5) How Much More Effective Is It to Test Yourself in Studying? We have...
Construct a 95% confidence interval of the population proportion using the given information. x= 125, n...
Construct a 95% confidence interval of the population proportion using the given information. x= 125, n = 250 Click here to view the table of critical values. The lower bound is The upper bound is (Round to three decimal places as needed.) i Table of critical values x Level of Confidence, (1 - «) - 100% CK Area in Each Tail, 2 Critical Value, 2 90% 0.05 1.645 95% 0.025 1.96 2.575 99% 0.005 Print Done
9.1.15 Construct a 99% confidence interval of the population proportion using the given information. x =...
9.1.15 Construct a 99% confidence interval of the population proportion using the given information. x = 125, n = 250 Click here to view the table of critical values. The lower bound is The upper bound is (Round to three decimal places as needed.) Table of critical values -X Area in Each Tail, i Critical Value, Level of Confidence, (1 - a). 100% 90% 95% 99% 0.05 0.025 0.005 1645 1.96 2.575 Print Done Enter your answer in the edit...
Name Date Points indicated in l's Confidence Interval Activities 1. a. Draw the standard normal curve...
Name Date Points indicated in l's Confidence Interval Activities 1. a. Draw the standard normal curve and shade the area corresponding to the middle 90% of the standard normal distribution. Determine the values-z and z such that the area under the standard normal curve between-z and z is equal to 0.90. Round the z-values to the nearest thousandth. [(4) אto ,י +0100enor m C.9s,0,1)=045 .go .oS 210.5.0 -\.045 -2 b. Repeat part (a) for the middle 95% of the distribution....
The distribution was created
using 1000 bootstrap statistics. Use the distribution to estimate
a? 99% confidence interval for the mean IQ for the population.
Round your answers to one decimal place.
The 99% confidence interval is ______ to ______.
Note - Please solve it for 99% CI, not 95%
CI.
IQ Scores A sample of 10 IQ scores was used to create the bootstrap distribution of sample means in Figure 1 # sample 1000 mean 100.104 st.dev. 4.798 80 |...
Here are the percentiles for a bootstrap distribution of average NHL Penalty minutes Percentiles for a bootstrap distribution of penalty minutes 0.50% 1.00% 2.00% 2.50% 5.00% Percentile 21.8 22.8 23.9 24.4 25.8 95.00% 99.50% 97.50% 45.5 98.00% 46.1 99.00% 47.7 43.5 49.5 1. Construct a 99% Confidence interval for the true mean Penalty Minutes. 2. Construct a 98% Confidence interval for the true mean Penalty Minutes. 3. Construct a 96% Confidence interval for the true mean Penalty Minutes. 4. Construct...
Find the indicated confidence interval. Assume the standard error comes from a bootstrap distribution that is approximately normally distributed. A 90% confidence interval for a mean if the sample has n = 100 with à = 22.4 and s = 5.7 , and the standard error is SE = 0.57 Round your answers to three decimal places. The 90% confidence interval is
Construct a 95% confidence interval of the population proportion using the given information. x= 125, n = 250 Click here to view the table of critical values. The lower bound is The upper bound is (Round to three decimal places as needed.) i Table of critical values x Level of Confidence, (1 - «) - 100% CK Area in Each Tail, 2 Critical Value, 2 90% 0.05 1.645 95% 0.025 1.96 2.575 99% 0.005 Print Done
9.1.15 Construct a 99% confidence interval of the population proportion using the given information. x = 125, n = 250 Click here to view the table of critical values. The lower bound is The upper bound is (Round to three decimal places as needed.) Table of critical values -X Area in Each Tail, i Critical Value, Level of Confidence, (1 - a). 100% 90% 95% 99% 0.05 0.025 0.005 1645 1.96 2.575 Print Done Enter your answer in the edit...
Name Date Points indicated in l's Confidence Interval Activities 1. a. Draw the standard normal curve and shade the area corresponding to the middle 90% of the standard normal distribution. Determine the values-z and z such that the area under the standard normal curve between-z and z is equal to 0.90. Round the z-values to the nearest thousandth. [(4) אto ,י +0100enor m C.9s,0,1)=045 .go .oS 210.5.0 -\.045 -2 b. Repeat part (a) for the middle 95% of the distribution....
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