Answer:
40 ounces of coffee
Step-by-step explanation:
we know that
To make caffe mocha, LaVelle uses one ounce of chocolate syrup,and five ounces of coffee
so
To make 6 ounces of caffe mocha (one ounce of chocolate syrup plus 5 ounces of coffe), LaVelle uses one ounce of chocolate syrup,and five ounces of coffee
using proportion
Find out how many ounces of coffee does she need to make 48 ounces of caffe mocha
Let
x ---> ounces of coffee needed to make 48 ounces of caffe mocha
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><em> = true mean transmission of pages.</em>
So, Null Hypothesis, :
10 pages {means that the true mean is smaller than or equal to 10}
Alternate Hypothesis, :
> 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. = ~
where, = 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> = ~
= 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( > 5.9028) = Less than 0.05% {using t table)