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A study is conducted to determine if a newly designed text book is more helpful to learning the material than the old edition. the mean score
$5.00$3.00Question: a study is conducted to determine if a newly designed text book is more helpful to learning the material than the old edition. the mean score on the final exam for a course using the old edition is 75. ten randomly selected people who used the new text take the final exam their scores are shown.
person: A,B,C,D,E,F,G,H,I,J
score: 90,74,96,95,89,71,80,86,67,78
Use a 0.01significance level to test the claim that people do better with the new edition. Assume the standard deviation is 10.5. ( Note you may wish to use statistical software).
A.)what kind of test should be used?
1. twosided tail
2. one tailed
3. it does not matter
B.) the test statistic is (rounded to two decimals)
C.) the Pvalue is
D.) is there sufficient evidence to support the claim that people do better than 75 on this exam? Y/N
E.) construct a 99% confidence interval for the mean score for students using the new text. ____<mean<____ 
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A study of the ability of individuals to walk in a straight line (” Can we Really walk straight?” Amer. J. of Physical Anthro., 1992: 1927)
$3.00$2.00 
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Suppose a baker claims that the average bread height is more than 15cm. Several of this customers do not believe
$3.00$2.00Question: Suppose a baker claims that the average bread height is more than 15cm. Several of this customers do not believe him. To persuade his customers that he is right, the baker decides to do a hypothesis test. He bakes 10 loaves of bread. The mean height of the sample loaves is 17 cm with a sample standard deviation of 1.9 cm. The heights of all bread loaves are assumed to be normally distributed. The baker is now interested in obtaining a 95% confidence interval for the true mean height of his loaves. What is the lower bound to this confidence interval? cm (round to 2 decimal places) What is the upper bound to this confidence interval? decimal places) cm (round to 2 decimal places)

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A manufacturer of pharmaceutical products analyses each batch of a product to verify the concentration of the active ingredient. The chemical
$2.00$1.00Question: confidence interval for mean
A manufacturer of pharmaceutical products analyses each batch of a product to verify the concentration of the active ingredient. The chemical analysis is not perfectly precise. In fact, repeated measurements follow a Normal distribution with mean μ equal to the true concentration and standard deviation σ = 0.0076 grams per litter. Three analyses of one batch give concentrations 0.8412, 0.8307, and 0.8321 g/l. To estimate the true concentration, give a 95% confidence interval for μ. (Round your answers to four decimal places.)

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Based on a random sample of 1040 adults, the mean amount of sleep per night is 7.73 hours
$2.00$1.00Question:
Based on a random sample of 1040 adults, the mean amount of sleep per night is 7.73 hours. Assuming the population standard deviation for amount of sleep per night is 3.5 hours. Construct and interpret a 90% confidence interval for the mean amount of sleep per night.
A 90% confidence interval is ( _____ , ______ ) 
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The life in hours of an electronic sensor is known to be approximately Normally distributed
$3.00$2.00Question:
The life in hours of an electronic sensor is known to be approximately Normally distributed, with population standard deviation = hours. A random sample of 10 sensors resulted in the following data: 500,550,560,575,525,505,510,540,550,545.
 Is there evidence to support the claim that mean life exceeds 520 hours? Use a fixedlevel test with alpha = 0.05.
 What is the Pvalue of this test? Conclude the same of part (a)?
 What is the Betavalue for this test if the true mean life is 535 hours?
 What sample size would be required to ensure that Beta does not exceed 0.10 if the true mean life is 540 hours?
 Construct a 95% onesided lower CI on the mean life.
 Use the CI found in part e) to test the hypothesis.

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A random sample of the number of farms (in thousands) in various states follows. Estimate the mean number of farms per state with 90 % confidence
$2.00$1.00Question:
A random sample of the number of farms (in thousands) in various states follows. Estimate the mean number of farms per state with 90 % confidence.
a) Assume population standard deviation is known and is equal to 31.
b) Assume that population standard deviation is not known.
47 95 54 33 64 48 57 9 80 8 90 3 49 4 44 79 80 48 16 68 7 15 21 52 6 78 109 40 50 29 5.

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A different sample of size n = 20 produced a sample mean of 31.08% and a sample standard deviation of 5.48%.
$2.00$1.00Question:
A different sample of size n = 20 produced a sample mean of 31.08% and a sample standard deviation of 5.48%. Use these values to calculate a 90% confidence interval for the national mean percent of lowincome working families. Please show all calculations and provide the upper and lower limits that make up the confidence interval. (Round the limits to two decimal places).