Answer:
We conclude that a web-based company are not exceeding their goal of 90%.
Step-by-step explanation:
We are given that a web-based company has a goal of processing 90 percent of its orders on the same day they are received.
434 out of the next 471 orders are processed on the same day.
Let p = <u><em>proportion of orders processing on the same day they are received.</em></u>
SO, Null Hypothesis, 
: p 
0.90 {means that they are not exceeding their goal of 90%}
Alternate Hypothesis, : p > 0.90 {means that they are exceeding their goal of 90%}
The test statistics that would be used here <u>One-sample z test for</u> <u>proportions</u>;
T.S. = 
~ N(0,1)
where, 
= sample proportion of orders that are processed on the same day = 
= 0.92
n = sample of orders = 471
So, <u><em>the test statistics</em></u> = 
= 1.599
The value of z test statistics is 1.599.
<u>Now, at 0.025 significance level the z table gives critical value of 1.96 for right-tailed test.</u>
Since our test statistic is less than the critical value of z as 1.599 < 1.96, so we have insufficient evidence to reject our null hypothesis as it will not fall in the rejection region due to which <u><em>we fail to reject our null hypothesis</em></u>.
Therefore, we conclude that a web-based company are not exceeding their goal of 90%.
1) The outcomes for rolling two dice, the sample space, is as follows:
(1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (1, 6)
(2, 1), (2, 2), (2, 3), (2, 4), (2, 5), (2, 6)
(3, 1), (3, 2), (3, 3), (3, 4), (3, 5), (3, 6)
(4, 1), (4, 2), (4, 3), (4, 4), (4, 5), (4, 6)
(5, 1), (5, 2), (5, 3), (5, 4), (5, 5), (5, 6)
(6, 1), (6, 2), (6, 3), (6, 4), (6, 5), (6, 6)
There are 36 outcomes in the sample space.
2) The ways to roll an odd sum when rolling two dice are:
(1, 2), (1, 4), (1, 6), (2, 1), (2, 3), (2, 5), (3, 2), (3, 4), (3, 6), (4, 1), (4, 3), (4, 5), (5, 2), (5, 4), (5, 6), (6, 1), (6, 3), (6, 5). There are 18 outcomes in this event.
3) The probability of rolling an odd sum is 18/36 = 1/2 = 0.5
That's a funky problem... :/ I mean it would depend on how much she earns weekly. If she were working 40 hours each week and earning 10$ an hour then yes, she would have enough. Even is she were per say a student on a part time working 30 hours and earning 8$ per hour, she would still have enough.
