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Svetradugi [14.3K]
3 years ago
9

Faced with rising fax costs, a firm issued a guideline that transmissions of 10 pages or more should be sent by 2-day mail inste

ad. Exceptions are allowed, but they want the average to be 10 or below. 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. (a-1) Find the test statistic. (Round your answer to 4 decimal places.) The test statistic (a-2) At the .01 level of significance, is the true mean greater than 10? No Yes (b) Use Excel to find the right-tail p-value. p-value
Mathematics
1 answer:
Vladimir [108]3 years ago
7 0

Answer:

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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