The nicotine content in cigarettes of a certain brand is normally distributed with mean (in milligrams) u and standard deviation a = 0.1. The brand advertises that the mean nicotine content of their Cigarettes is 1.5 mg. Now, suppose a reporter wants to test whether the mean nicotine content is actually higher than advertised, He takes measurements from a SRS of 15 cigarettes of this brand. The sample yields an average of 1.45 mg of nicotine. Conduct a test using a significance level of a 0.05. (a) The test statistic (b) The critical value, z* (c) The final conclusion is A. The nicotine content is probably higher than advertised. B. There is not sufficient evidence show that the ad is misleading.

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A: (a)   Obtain the probability that X is more than 60. The probability that is more than 60 is…

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A:

Q: The frequency distribution was obtained using a class width of 0.5 for data on cigarette tax rates.…

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Q: The frequency distribution was obtained using a class width of 0.5 for data on cigarette tax rates.…

A: Frequency distribution of cigarettes tax rate. We have to find out population mean and population…

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A: Solution: Before After d = Before - After  (d-d) (d-d)2 39.2 40 -0.8 -1.5 2.25 38.5 37.9 0.6…

Q: The frequency distribution was obtained using a class width of 0.5 for data on cigarette tax rates.…

A:

Q: A person's blood glucose level and diabetes are closely related. Let x be a random variable measured…

A: NORMAL DISTRIBUTION:- Normal distribution is a continues distribution of data that has the…

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A:

Q: A person's blood glucose level and diabetes are closely related. Let x be a random variable measured…

A: Since you have posted multiple sub part ,we are only provide the answer of forst three subpart…

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A: From the provided information, The height of young men are approximately normal with standard…

Q: A person's blood glucose level and diabetes are closely related. Let x be a random variable measured…

A: Given information: mean, μ=86 Standard deviation, σ=29. a) The probability that x is more than 60…

Q: A simple random sample of six values had mean x = 13.00 and standard deviation s = 7.155. Calculate…

A: Given,

Q: Blood pressure is distributed Normally, with a mean of 85 mm and a standard deviation of 20 mm.…

A:

Q: A population of scores has μ = 80. In this population, a score of X= 92 corresponds to z = 5. what…

A: Given: μ = 80 z = 5 X= 92

Q: A person's blood glucose level and diabetes are closely related. Let x be a random variable measured…

A: Normal distribution is a continuous distribution that has many real-life applications. A normal…

Q: A person's blood glucose level and diabetes are closely related. Let x be a random variable measured…

A: GivenMean(μ)=84standard deviation(σ)=21

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A: Solution : Given : μ = 120.1             σ = 15.1 Let , X : Systolic blood pressures Here, X ~ N (…

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A: The sample size is 31, sample mean is 9.1 and sample standard deviation is 5.3.

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A:

Q: The standard deviation of pulse rates of adult males is  less 11 bpm. For a random sample of  128…

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Q: Blood pressure is distributed Normally, with a mean of 85 mm and a standard deviation of 20 mm.…

A: The objective of this question is to find the fraction of the population that has a blood pressure…

Q: deviation s = 0.1. Is there sufficient evidence in the sample to suggest that the mean nicotine…

A: here given, sample mean  = 1.53 sample standard deviation  = s= 0.1 significance level = 0.01

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A: Given data, Mean=25 sd =2.5 If n=9 P(23<X<26)=?

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A:

Q: The degenerative disease osteoarthritis most frequently affects weight-bearing joints such as the…

A: Note: According to the guidelines, since there are more than 4 subparts, we will answer the first…

Estimation

A new framework of data analysis where the new statistical methods are used for interpreting the results and analyzing the data is known as estimation in statistics.