A large home improvement store is considering expanding its selection of moving products, such as cardboard boxes and packing tape. The store constructs a 90% confidence interval to estimate the proportion of all customers
who have moved houses at least once in the last five years. In the random sample of 620 customers, 201 (32.4%)
replied that they had moved at least once in the last five years. The sample yielded the 90% confidence interval
(0.293, 0.355) for the proportion of all customers who have moved in the past five years. Sheldon, an employee who
wishes to expand the selection of moving products, claims that more than 1 in 4 customers have moved in the past
five years. Zachary, another employee, urges Sheldon to be bolder in his claim and say that more than 1 in 3
customers have moved in the past five years. Based on the confidence interval, which employee's claim is
plausible?

Based on the confidence interval, Sheldon's claim is plausible

Step-by-step explanation:

The number of customers in the sample, n = 620 customers

The number of customers that replied that they had moved at least once in the last five years = 201 (32.4%)

The 90% confidence interval for the sample = (0.293, 0.355)

Let 'p' represent the proportion of all customers who have moved in the past five years

The proportion of customers Sheldon claimed have moved in the past five years, p > 1 in 4 (0.25)

The proportion Zachary urges Sheldon to claim is, p > 1 in 3 (0.)

Given that the lower estimate of the confidence interval is 0.293 (>0.25) and upper bound of the confidence interval is 0.355 (> 0.), and that 0.26 in also more than 1 in 4 but less than 0.293, but however that 0.293 which is within the 90% confidence interval is less than 0. or 1 in 3, as opposed to what Zachary urges Sheldon to claim, Sheldon's claim is more plausible.

step-by-step explanation:

i found the life tables for the year 2012. 

life table of males at age 30. probability of dying between ages x and x+1

total population:  

29–30 . . . . . . . . . . . . 0.001030

30–31 . . . . . . . . . . . . 0.001061

males:  

29–30 . . . . . . . . . . . . 0.001420

30–31 . . . . . . . . . . . . 0.001448

white population

29–30 . . . . . . . . . . . . 0.001005

30–31 . . . . . . . . . . . . 0.001037

white males

29–30 . . . . . . . . . . . . 0.001372

30–31 . . . . . . . . . . . . 0.001399

non hispanic white population

29–30 . . . . . . . . . . . . 0.001083

30–31 . . . . . . . . . . . . 0.001116

non hispanic white males

29–30 . . . . . . . . . . . . 0.001477

30–31 . . . . . . . . . . . . 0.001503

jacob, non hispanic white male. 30 years old.

probability that jacob lives to be 30 years old. 

since the given probabilities are probabilities of dying, i think we simply subtract it from 1 to get the probability of living. we will use the probability of non hispanic white male.

                                          probability of living to be 30 y.o.

29 - 30 : 1 - 0.001477 = 0.998523

30 - 31 : 1 - 0.001503 = 0.998497

id

step-by-step explanation:

idk

idk

i know the way

and

idk