Question

The effectiveness of a blood-pressure drug is being investigated. An experimenter finds that, on average, the reduction in systolic blood pressure is 28.1 for a sample of size 23 and standard deviation 20.3. Estimate how much the drug will lower a typical patient's systolic blood pressure (using a 90% confidence level). Assume the data is from a normally distributed population.

Enter your answer as a tri-linear inequality accurate to three decimal places.

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Karen wants to advertise how many chocolate chips are in each Big Chip cookie at her bakery. She randomly selects a sample of 45 cookies and finds that the number of chocolate chips per cookie in the sample has a mean of 5.1 and a standard deviation of 3.1. What is the 99% confidence interval for the number of chocolate chips per cookie for Big Chip cookies? Enter your answers accurate to one decimal place (because the sample statistics are reported accurate to one decimal place).

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Karen wants to advertise how many chocolate chips are in each Big Chip cookie at her bakery. She randomly selects a sample of 45 cookies and finds that the number of chocolate chips per cookie in the sample has a mean of 5.1 and a standard deviation of 3.1. What is the 99% confidence interval for the number of chocolate chips per cookie for Big Chip cookies? Enter your answers accurate to one decimal place (because the sample statistics are reported accurate to one decimal place).

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Answer 1

(1)For the effectiveness of a blood-pressure drug being investigated given data is taken

Sample Size = n = 23

Sample mean = X = 28.1

Saple standard deviation = s = 20.3

For 90 % confidence, level of significance α=0.1

Degree of freedom = sample size - 1 = 23-1 = 22

For degree of freedom 22 and α=0.1 critical t value using excel formula =T.INV.2T(0.1,22) (Note - we are using absolute value of t here,otherwise for a two tailed test we get two values of t ,one in negative sign with same magnitude and other in positive)

tc = 1.717

Now confidence interval is calculated using

90 % confidence interval for the population mean μ is

(2)Karen randomly selects a sample of 45 cookies and finds that the number of chocolate chips per cookie in the sample has a mean of 5.1 and a standard deviation of 3.1. So given data is

Sample Size = n = 45

Sample mean = X = 5.1

Saple standard deviation = s = 3.1

For 99 % confidence ,level of significance α=0.01

Degree of freedom = sample size - 1 = 45-1 = 44

For degree of freedom 44 and α=0.01 critical t value using excel formula =T.INV.2T(0.01,44)

tc = 2.692

99 % confidence interval for the population mean μ is

3rd question is same as 2nd question , so solution will be same