Find the P-value for a left-tailed hypothesis test with a test statistic of z= -1.55. Decide whether to reject Upper H 0H0 if
1 answer:
Answer:
0.0606
Step-by-step explanation:
Given that test statistic for a left tailed test is
z=-1.55
Significance level = 0.10
let us find p value from z value
p value for left tailed is
0.060571
Since p value < 0.10 our alpha, we reject null hypothesis.
p value = 0.0606
Conclusion: Reject null hypothesis.
There is statistical evidence to show that sample mean is less than population mean.
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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%
