The probability that a child with a speaking part is chosen randomly would be 2:5.
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.
The difference in the mean rainfall is given by 19.7 - 19.3 = 0.5
The difference in the <span>mean rainfall is approximately 0.5/4.6 = 0.087 = 8.7 percent of the mean absolute deviation for last week and is 0.5/5.2 = 0.077 = 7.7 percent of the mean absolute deviation for this week.
Therefore, the difference in the </span><span>mean rainfall is approximately what percent of the mean absolute deviation 8% of the mean absolute deviation of the data sets.</span>
10-3 or 10+(-3) that should help
Fifty-two billion, six hundred and thirty-four million, two hundred and seventy-five thousand, three hundred and nine.
