A bank wants to estimate the difference in the proportion of its "big" customers who have a second account with a different bank and the proportion of its "small" customers who also have an account with another bank. Of the 200 "big" customers randomly selected, 160 claimed to also bank elsewhere. Of the 300 "small" customers randomly selected, 120 claimed to also bank with other banks. Check the conditions for constructing a confidence interval for the difference in proportions.

Kindly check explanation

Step-by-step explanation:

To construct confidence interval for difference in proportion ; the following conditions must be met :

1.) The selected samples must have been randomly chosen or selected from a larger population. This is essential in other to eliminate bias.

Each group of samples must be independently chosen. That is, the samples of bug customers chosen must be independent of the sample of small customers selected.

The population from which the samples are drawn must be considerably larger than the sample size. Sample size, n should be ≤ 10% of the population size.

p1 = 160 / 200 = 0.8 ; (1 - p1) = 0.2

p2 = 120 / 300 = 0.4 ; (1 - p2) = 0.6

(p1 - p2) ± Zcritical * sqrt[(p1(1 - p1) /n1) + (p2(1 - p2) / n2)]

d. millimeters per hour.

step-by-step explanation:

we are given that,

carlo used a rain gauge to measure the amount of water his lawn is receiving from the sprinklers.

since, according to the options, we see that,

water cannot be measured by 'miles per hour', 'meters per second' and 'grams per seconds'.

we get that, the water can be measured by 'millimeters per hour'.

hence, the appropriate rate for describing the amount of water is 'millimeters per hour'.