Please read questions carefully and label all parts and their answers clearly Pa

Question

Please read questions carefully and label all parts and their answers clearly
Part 1
A pharmaceutical company investigating whether drug stores are less likely than food markets to remove over-the-counter drugs from the shelves when the drugs are past the expiration date found a P-value of 2.8%. This means that:

A) 2.8% more drug stores remove over-the-counter drugs from the shelves when the drugs are past the expiration date.
B) 97.2% more drug stores remove over-the-counter drugs from the shelves when the drugs are past the expiration date than drug stores.
C) none of these
D) There is a 2.8% chance the drug stores remove more expired over-the-counter drugs.
E) There is a 97.2% chance the drug stores remove more expired over-the-counter drugs.
Part 2 A newspaper article reported that a poll based on a sample of 1150 residents of a state showed that the state's Governor’s job approval rating stood at 58%. They claimed a margin of error of ±3%. What level of confidence were the pollsters using? Show all work using the equation editor beginning with the margin of error formula. Round your solution to the nearest whole percent. Write a sentence that gives your solution.
Please read questions carefully and label all parts and their answers clearly
Part 1
A pharmaceutical company investigating whether drug stores are less likely than food markets to remove over-the-counter drugs from the shelves when the drugs are past the expiration date found a P-value of 2.8%. This means that:

A) 2.8% more drug stores remove over-the-counter drugs from the shelves when the drugs are past the expiration date.
B) 97.2% more drug stores remove over-the-counter drugs from the shelves when the drugs are past the expiration date than drug stores.
C) none of these
D) There is a 2.8% chance the drug stores remove more expired over-the-counter drugs.
E) There is a 97.2% chance the drug stores remove more expired over-the-counter drugs.
Part 2 A newspaper article reported that a poll based on a sample of 1150 residents of a state showed that the state's Governor’s job approval rating stood at 58%. They claimed a margin of error of ±3%. What level of confidence were the pollsters using? Show all work using the equation editor beginning with the margin of error formula. Round your solution to the nearest whole percent. Write a sentence that gives your solution.
Please read questions carefully and label all parts and their answers clearly
Part 1
A pharmaceutical company investigating whether drug stores are less likely than food markets to remove over-the-counter drugs from the shelves when the drugs are past the expiration date found a P-value of 2.8%. This means that:

A) 2.8% more drug stores remove over-the-counter drugs from the shelves when the drugs are past the expiration date.
B) 97.2% more drug stores remove over-the-counter drugs from the shelves when the drugs are past the expiration date than drug stores.
C) none of these
D) There is a 2.8% chance the drug stores remove more expired over-the-counter drugs.
E) There is a 97.2% chance the drug stores remove more expired over-the-counter drugs.
Part 2 A newspaper article reported that a poll based on a sample of 1150 residents of a state showed that the state's Governor’s job approval rating stood at 58%. They claimed a margin of error of ±3%. What level of confidence were the pollsters using? Show all work using the equation editor beginning with the margin of error formula. Round your solution to the nearest whole percent. Write a sentence that gives your solution.

Explanation / Answer

Part 1) Solution: 2.8% = 0.028< 0.05(Level of significance)

We have not given null hypothesis therefore,

Option C) is correct.

Part 2) Solution: We have given,

p = 0.58
Standard error of p = sqrt(0.58)(0.42)/1150) =0.01455425

Margin of error =E = z (0.01455425), where z is a standard normal variate.
z (0.01455425) = 0.03
z = 0.03 / 0.01455425 = 2.06

Now, by using normal probability tables, P( z > 2.06)+P( z < -2.06) = 0.0197+0.0197 = 0.0394 = 0.04

Therefore, this is a 1-0.04 = .96 = 96% confidence interval.

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