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RoseWind [281]
1 year ago
11

Find the quotient 3^15/3^3

Mathematics
1 answer:
Sati [7]1 year ago
6 0

There is a property called "Quotient of powers property", which states that:

\frac{a^m}{a^n}=a^{(m-n)}

Where "a" is the common base and "m" and "n" are exponents.

For this case, you have:

\frac{3^{(15)}}{3^3}

Then, in order to find the quotient, you must apply the Quotient of powers property. You need to write the common base (in this case is 3) and then subtract the exponents.

So, you get that the quotient is:

\frac{3^{(15)}}{3^3}=3^{(15-3)}=3^{(12)}

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The value of t test statistics is 5.9028.

We conclude that the true mean is greater than 10 at the .01 level of significance.

Step-by-step explanation:

We are given that a firm issued a guideline that transmissions of 10 pages or more should be sent by 2-day mail instead.

The firm examined 35 randomly chosen fax transmissions during the next year, yielding a sample mean of 14.44 with a standard deviation of 4.45 pages.

<em />

<em>Let </em>\mu<em> = true mean transmission of pages.</em>

So, Null Hypothesis, H_0 : \mu \leq 10 pages     {means that the true mean is smaller than or equal to 10}

Alternate Hypothesis, H_A : \mu > 10 pages     {means that the true mean is greater than 10}

The test statistics that would be used here <u>One-sample t test statistics</u> as we don't know about the population standard deviation;

                        T.S. =  \frac{\bar X-\mu}{\frac{s}{\sqrt{n} } }  ~ t_n_-_1

where, \bar X = sample mean = 14.44 pages

            s = sample standard deviation = 4.45 pages

            n = sample of fax transmissions = 35

So, <u><em>test statistics</em></u>  =  \frac{14.44-10}{\frac{4.45}{\sqrt{35} } }  ~ t_3_4  

                               =  5.9028

(a) The value of t test statistics is 5.9028.

Now, at 0.01 significance level the z table gives critical values of 2.441 at 34 degree of freedom for right-tailed test.

<em>Since our test statistics is more than the critical values of z as 5.9028 > 2.441, so we have sufficient evidence to reject our null hypothesis as it will fall in the rejection region due to which </em><em><u>we reject our null hypothesis.</u></em>

Therefore, we conclude that the true mean is greater than 10.

(b) Now, P-value of the test statistics is given by the following formula;

               P-value = P( t_3_4 > 5.9028) = Less than 0.05%    {using t table)

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