Suppose that a company claims that its air fresheners last 23 days, on average. you take many large samples, and find, each time, that the mean number of days of the sample is outside the 95% confidence interval. based on this information alone, which of the following is probably not the mean number of days that the company's air fresheners last?

a. 19
b. 21
c. 17
d. 23

Option D is the answer.

Explanation:

A company claims that its air fresheners last 23 days, on average. Many large samples are taken each time it is found that the mean number of days of the sample is outside the 95% confidence interval.

This implies:

So, below 21.85 is the probable mean number of days that the company's air fresheners last.

Therefore, 23 is not the mean number of days that the company's air fresheners last.

many reason,

step-by-step explanation:

1. sometimes

2.   never

3   .08c = 7.50+.05c   250 pages

step-by-step explanation:

1.   sometimes

the theoretical probability of rolling a 1 is 1/6.   perhaps you got really lucky and never rolled a 1   in those 60 times.   your experimental probability is 0.   or perhaps you rolled a 1   20 times.   your experimental probability is 20/60 is 1/3.     it really just depends on how you rolled in that experiment.

2.   never

the probability of the event occurring is 1.   that is being 100 % sure it will happen.   that is the maximum amount of confidence we have.

3.   no customer card cost:   .08 times number of pages

customer card cost:   7.50 + .05 times number of pages

we want to find when it costs the same amount, so we will set these equal

let c = number of pages

.08c = 7.50+.05c

subtract .05c from each side

.08c-.05c = 7.5+.05c-.05c

.03c=7.5

divide each side by .03

.03c/.03 = 7.5/03

c =250  

the cost is the same at 250 pages