Answer:
-4.42
Step-by-step explanation:
Given the following :
Population mean (pm) = 8 hours
Sample mean (sm) = 6.9 hours
Number of observations (n) = 101 students
Sample standard deviation (sd) = 2.5
Using the t-statistic formula :
[(sm - pm) / (sd/√n)]
Hence,
[(6.9 - 8) / (2.5/√101)]
[(-1.1) / (2.5/10.05)]
-1.1 / 0.2487562
= −4.422
The test statistic = - 4.42 to the nearest hundredth.
Answer:
66
Step-by-step explanation:
Answer:
The correct answer is
(0.0128, 0.0532)
Step-by-step explanation:
In a sample with a number n of people surveyed with a probability of a success of
, and a confidence interval
, we have the following confidence interval of proportions.

In which
Z is the zscore that has a pvalue of 
For this problem, we have that:
In a random sample of 300 circuits, 10 are defective. This means that
and 
Calculate a 95% two-sided confidence interval on the fraction of defective circuits produced by this particular tool.
So
= 0.05, z is the value of Z that has a pvalue of
, so
.
The lower limit of this interval is:

The upper limit of this interval is:

The correct answer is
(0.0128, 0.0532)
She saved up a total of $82.60, and had to spend a total of $60.90, so she would have $21.70 left and need to save up another $180.40, hope this helps