Answer:31) The Coca-Cola Company reported that the mean per capita annual sales of its beverages in the United Sates was 423 eight-ounce servings. Suppose you are curious whether the consumption of Coca-Cola beverages is higher in Atlanta. A sample of 36 individuals from the Atlanta area showed a sample mean annual consumption of 460.4 eight ounce servings with a standard deviation of s=101.9 ounces. Using a=.05, do the sample results support the conclusion that mean annual consumption of Coca-Cola beverage products is higher in Atlanta
Step-by-step explanation:29) The national mean annual salary for a school administrator is $90,000 a year. (The Cincinnati Enquirer, April 7, 2012) A school official took a sample of 25 school administrators in the state of Ohio to learn about salaries in that state to see if they differed from the national average.
a) Formulate hypotheses that can be used to determine whether the population mean annaual administrator salary in Ohio differs from the nation mean of $90,000.
b) The sample data for 25 Ohio administrators is contained in the file named Administrator. What is the p-value for you hypothesis test in part (a)?
c) A a=.05 can your null hypothesis be rejected? What is your conclusion?
d) Repeat the preceding hypothesis test using the critical value approach.
Answer:
4
Step-by-step explanation:
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: 2%, second option is correct.
Step-by-step explanation:
To state 1/50 in percent, divide 1 by 50, then multiply by 100
=( 1 ÷ 50) x 100
= 0.02 x 100
= 2%
I hope this helps, please mark as brainliest.
Consider this option:
C³₂₇=27!/(3!*24!)=25*13*9=2925 ways to select 3 students.
