W/15 = 3/15
cross multiply 15 by 3
15 x 3 = 45
then divide 45 by 15
45 x 15 = 3
therfore W = 3
cross multiply 15 by 3
15 x 3 = 45
then divide 45 by 15
45 x 15 = 3
therfore W = 3
You need to earn at least $100. You earn $8.00 per hour mowing lawns. Write and solve an inequality that represents the number o
1 answer:
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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)