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Confidence interval for mean to assess the risks

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1. To help consumers assess the risks they are taking, the food and drug administration publishes the amount of nicotine found in all commercial brands of cigarettes. A new cigarette has recently been marketed. The FDA tests on this cigarette gave a mean nicotine of content 28.2 milligrams and standard deviation of 2.7 milligrams for a sample of n = 9. The FDA claims that the mean nicotine content exceeds 32.2 milligrams for this brand of cigarette, and their stated reliability is 99%. Do you agree

1. no, since the value 32.2 does fall in the 99% confidence interval

2. yes, since the value 32.2 does not fall in the 99% confidence interval

3.no since the value 32.2 does not fall in the 99% confidence interval

4. yes, since the value 32.2 does fall in the 99% confidence interval

2. A confidence interval was used to estimate the proportion of statistics students that are female. A random sample of 72 statistics students generated the following 99% confidence interval. State the confidence level used to create the confidence interval.
1. between 43.8% and 64.2%
2. 99%
3. 64.2%
4. 72%

3. A survey claims that 9 out 0f 10 doctors (i.e 90%) recommend brand z for their children. To test this claim against the alternative that the actual proportion of doctors who recommend brand Z is less than 90%, a random sample of doctors was taken. Suppose the test statistic is z = -1.95. can we conclude that Ho should be rejected at the a) a = 0.10 b) a = 0.05 and C) 0.01 level?

a. a)no b)yes c)yes
b. a)yes b)yes c)no
c. a)yes b)yes c)yes
d. a)no b)no c)no

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The solution gives the details of construction of confidence interval for mean and proportion.

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