Answer my question and i will mark you as brainliest please help me its that easy
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Answer:
23/56 centimeters
Step-by-step explanation:
all of the variables have a common denominator, so you can effectively ignore the denominator when solving.
1. you start off with 34/56 centimeter of chalk and Brian breaks off 23/56 centimeters worth, leaving you with 11/56 centimeters
2. Sara then added 12/56 centimeters worth, making the final amount 23/56 centimeters of chalk
An equation to summarize all of this is:
Answer:
Check the explanations
Step-by-step explanation:
According to given information, an instructor at a major research university occasionally teaches summer session and notices that that there are often students repeating the class. On the first day of class, she counts 105 students enrolled, of which 19 are repeating the class. The university enrolls 15,000 students.
Therefore the number of observation n 105 and enrolled x=19
1. An estimate of the population proportion repeating the class is given by:
c) 0.181.
Explanation:
19 --0.18090,181 n 105
2. The instructor wishes to estimate the proportion of students across campus who repeat a course during summer sessions and decides to do so on the basis of this class. Would you advise the instructor against it and why:
d) Yes, because this class is not a random sample of students.
3. The standard error for the estimated sample proportion is given by:
c) 0.0376.
Explanation:
SE
0.181 × (1-0.181) 105
0,0376 0.0376
4. A 95% confidence interval is given by:
c) 0.107, 0.255.
Explanation:
In order to determine the 95% confidence interval we follow the following step:
Where the z value is determined from the standard normal table as ~ 1.96
0.181 ± (1.96 × 0.0376)
0.181 ± 0.0737
Therefore the lower confidence interval is
LCI= 0.181- 0.0737
LCI= 0.107
Therefore the upper confidence interval is
UCI = 0.181 + 0.0737
UCI0.255
Therefore 95% confidence interval is
5. She hypothesizes that, in general, 10% of students repeat a course. The hypotheses to be tested are:
d)
Explanation:
She hypothesizes that, in general, 10% of students repeat a course. The hypotheses to be tested are defined on the basis of observation,
Null hypothesis as:
:p= 0.10
and alternative hypothesis as:
6. The test statistic for this hypothesis is given by:
c) 2.765.
Explanation:
In order to determine the z test statistics as:
Z=
0.181 - 0.10 0.10x (1-0.10)
=2.765
7. The P value for this test is:
c) 0.01>P>0.005 .
Explanation:
P value is calculated as:
P(Z > 2.765) 0.002845 for one tail test and
P(Z > 2.765) 0.005692 for two tail test.
8. Based on the p-value found:
b) we have strong evidence that the proportion of students repeating a class during summer sessions is not 10%.
Explanation:
As the z observed is more than the tabulated z value at 95% as:
and also P value is less than the
Therefore we accept the alternative hypothesis and we may conclude that we have strong evidence that the proportion of students repeating a class during summer sessions is not 10%.