In order to study the amounts owed to a particular city, a city clerk takes a random sample of 16

files from a cabinet containing a large number of delinquent accounts and finds the average amount

rowed to the city to be $230 with a sample standard deviation of $36. It has been claimed that the

true mean amount owed on accounts of this type is greater than $250. If it is appropriate to assume

that the amount owed is a Normally distributed random variable, the value of the test statistic

appropriate for testing the claim is

(a) -3. 33 (b) - 1. 96

(c) - 2. 22

(d) -0. 55

(e) - 2. 1314

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joaobezerra joaobezerra · Answer 1 · 2022-04-28T11:05:41+02:00

Using the t-distribution, it is found that the value of the test statistic is given by:

(c) -2. 22

At the null hypotheses, it is tested if the mean is not greater than 250, that is:

[tex]H_0: \mu \leq 250[/tex]

At the alternative hypotheses, it is tested if it is greater, that is:

[tex]H_1: \mu > 250[/tex]

It is given by:

[tex]t = \frac{\overline{x} - \mu}{\frac{s}{\sqrt{n}}}[/tex]

The parameters are:

In this problem, the values of the parameters are:

[tex]\overline{x} = 230, \mu = 250, s = 36, n = 16[/tex].

Hence, the test statistic is given by:

[tex]t = \frac{\overline{x} - \mu}{\frac{s}{\sqrt{n}}}[/tex]

[tex]t = \frac{230 - 250}{\frac{36}{\sqrt{16}}}[/tex]

t = -2.22.

Which means that option C is correct.

More can be learned about the t-distribution at https://brainly.com/question/26454209