Answer:
it depends how much the cds cost.
Answer:
Step-by-step explanation:
We would set up the hypothesis test. This is a test of a single population mean since we are dealing with mean
For the null hypothesis,
µ = 17
For the alternative hypothesis,
µ < 17
This is a left tailed test.
Since the population standard deviation is not given, the distribution is a student's t.
Since n = 80,
Degrees of freedom, df = n - 1 = 80 - 1 = 79
t = (x - µ)/(s/√n)
Where
x = sample mean = 15.6
µ = population mean = 17
s = samples standard deviation = 4.5
t = (15.6 - 17)/(4.5/√80) = - 2.78
We would determine the p value using the t test calculator. It becomes
p = 0.0034
Since alpha, 0.05 > than the p value, 0.0043, then we would reject the null hypothesis.
The data supports the professor’s claim. The average number of hours per week spent studying for students at her college is less than 17 hours per week.
Answer:
12 the answer
Step-by-step explanation:
B. The median of the data set
**Update; A. The lower quartile**
See other answer for explanation.
1. centimeter
2. Feet
3. Miles
4. Feet
This is just my best guess
