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cupoosta [38]
3 years ago
9

It is recommended that adults get 8 hours of sleep each night. A researcher hypothesized college students got less than the reco

mmended number of hours of sleep each​ night, on average. The researcher randomly sampled 20 college students and found no evidence to reject the null hypothesis at the​ 5% significance level. What is true regarding the​ p-value from this hypothesis​ test?
Mathematics
1 answer:
yanalaym [24]3 years ago
5 0

Answer:

Given:

Sample size, n = 20

Significance level = 5% = 0.05

Mean, u = 8

Here, the sample mean was not given.

Here the null and alternative hypothesis will be:

H0 : u = 8

H1 : u < 8

In this case, since there is no evidence to reject the null hypothesis, H0, at 0.05 level of significance, we can say that the p-value is greater than the level of significance, 0.05.

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What is the length of AD if A is located at (-7,0) and D is located at (8,0)?
Kryger [21]

Answer:

15

Step-by-step explanation:

(-7,0) and (8,0) are on the opposite side of origin of x axis

d = 8 - -7 = 15

6 0
3 years ago
Find the equivalent expression 5^6
Naily [24]
5^12/5^6 Is the correct response. When equal numbers with different exponents are divided then the you must subtract the exponents. 
Thus 12-6 = 6. 
6 0
3 years ago
Sorting through unsolicited e-mail and spam affects the productivity of office workers. An InsightExpress survey monitored offic
nydimaria [60]

Answer:

1. Refer to the explanation for the frequency table.

2. (a) 60%

   (b) 25%

Step-by-step explanation:

Part 1. The question is asking us to group the data according to the classes given and fill in the frequency (F), relative frequency (RF), cumulative frequency (CF) and relative cumulative frequency (RCF).

To compute the frequency for each class count the number of data that lies in the class range and write it. For the first class (1-5) we can see that there are 12 numbers which lie in the range 1 to 5. Those numbers are: 2, 4, 4, 1, 2, 1, 5, 5, 5, 3, 4, 4. Similarly, for the second class, the frequency is 3 because 8, 8, 8 lie in this range from our existing data. For the third class 11-15 there are 2 numbers in the given data which lie in this range and those numbers are 12, 15. Similarly, the rest of the frequencies can be computed.

For the relative frequency, the frequency of each class is divided by the total frequency i.e. 20.  

For the class 1-5, RF = 12/20 = 0.6.  

For 6-10, RF = 3/20 = 0.15

For 11-15, RF = 2/20 = 0.1

For 16-20, RF = 1/20 = 0.05

For 21-25, RF=1/20 = 0.05

For 26-30, RF = 0/20 = 0.00

For 31-34, RF= 1/20 = 0.05.

To compute the cumulative frequency, add the existing frequency of each class with the previous frequencies.

For 1-5, CF = 0+12 = 12

For 6-10, CF = 12+3=15

For 11-15, CF = 12+3+2 = 17

For 16-20, CF=12+3+2+1 = 18

For 21-25, CF = 12+3+2+1+1=19

For 25-30, CF = 12+3+2+1+1+0=19

For 31-34, CF = 12+3+2+1+1+0+1=20

Now, to compute relative cumulative frequency, add the existing relative frequency of each class with the previous relative frequencies.  

For 1-5, RCF = 0+0.6

For 6-10, RCF = 0.6 + 0.15 = 0.75

For 11-15, RCF = 0.6+0.15+0.1 = 0.85

The rest of the relative cumulative frequencies can be computed in the same way.

Class(Minutes)           F       RF     CF     RCF

      1-5                       12     0.60    12     0.60  

      6-10                      3     0.15     15     0.75  

      11-15                      2     0.10     17     0.85  

      16-20                    1      0.05    18     0.90  

      21-25                    1      0.05    19     0.95  

      26-30                   0     0.00    19     0.95  

      31-34                     1     0.05    20       1  

       <u>Total                  20</u>  

Part 2. Now, we are asked to compute the Ogive graph which is also called as the cumulative frequency graph. The cumulative frequency needs to be plotted on the y-axis and the upper limit of each class needs to be plotted on the x-axis. The graph is attached.

(a) From the graph we can see that the number of workers who spend less than 5 minutes on unsolicited e-mail and spam are 12. So the answer for this part is 12 workers.

Percentage = 12/20 x 100 = 60%

(b) From the graph we can see that the number of workers who spend less than 10 minutes on spam e-mail are 15. The question is asking for the number of people who spend more than 10 minutes. For this we need to subtract 15 from the total number of workers.  

Number of workers spending more than 10 minutes = 20-15 = 5 workers.

Percentage = 5/20 x 100 = 25%

4 0
3 years ago
Solve for y py+qy=-4y+8
Dahasolnce [82]

Answer:

y = 8/p+q+4.

Step-by-step explanation:

5 0
3 years ago
An airplane took 5 hours to travel 3,420 kilometers.how many kilometers did the plane travel in one hour
tino4ka555 [31]

Answer:

684 Hours.

Step-by-step explanation:

If the plane travelled that distance in 5 hours, then all you need to do is divide the distance by 5 to get the distance travelled in one hour. This is obviously presuming that the speed travelled was constant or that this is an average figure.

3420/5 = 684.

8 0
4 years ago
Read 2 more answers
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