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Lina20 [59]
3 years ago
12

Consider a binomial experiment with n trials and r successes. To construct a test for a proportion p, what value do we use for t

he z value of the sample test statistic?
r/n
(r/n)2
n/r
2r/n
Mathematics
1 answer:
madam [21]3 years ago
3 0

Answer:

value we use for z value of  sample test  is r/n

Step-by-step explanation:

Given data

proportion p

trials  = n

successes = r

to find out

what value we use for z value of  sample test

solution

here

we know that

Probability of a success in single trail test = Number of successes/Number of trails

Probability of a success in single trail = r / n

so that

value we use for z value of  sample test  is r / n

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Step-by-step explanation:

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11111nata11111 [884]

Answer:

The most common salary is $605. The salary that half the employees salaries surpass is $630. The percent of employees salary that serve because $700 is 75%. Percent of employees salaries that were less than $480 is 25%. The 9% percent of employees salaries this surpass $891 years. The total weekly salary of 104 employees is 57200.

Step-by-step explanation:

It is given that mean is $550, first quartile is $480,  median is $630, third quartile is $700,  mode is $605 and 91st percentile is $891.

First quartile is at 25% of the data, median is at 50% of the data, third quartile is at 75% of the data.

The mode of a set of data values is the value that occurs most often.

The most common salary is mode, therefore the most common salary is $605.

The salary that half the employees salaries surpass is $630. Because median is the half of the data.

The percent of employees salary that serve because $700 is 75%. Because $700 is third quartile.

Percent of employees salaries that were less than $480 is 25%. Because $480 is first quartile.

The percent of employees salaries this surpass $891 years is 9%. Because $891 is 91 percentile. Therefore 9% employees get more than $891.

If the company has 104 employees than the total salary of employees is

104\times 550=57200

because means is $550, therefore the total weekly salary of 104 employees is 57200.

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