Question

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

Answer 1

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.

Let $\mu$ = true mean transmission of pages.

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 One-sample t test statistics 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, test statistics  =  $\frac{14.44-10}{\frac{4.45}{\sqrt{35} } }$  ~ $t_3_4$  

                               =  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.

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 we reject our null hypothesis.

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)