Answer:
Since the null hypothesis is true, finding the significance is a type I error.
The probability of the year I error = level of significance = 0.05.
so, the number of tests that will be incorrectly found significant is computed as follow: 0.05 * 100 = 5
Therefore, 5 tests will be incorrectly found significant given that the null hypothesis is true.
Answer:
73.6
Step-by-step explanation:
64x15%=9.6
9.6+64=73.6
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Step-by-step explanation:
Solution :
Given :
Sample mean, 
Sample size, n = 129
Sample standard deviation, s = 8.2
a. Since the population standard deviation is unknown, therefore, we use the t-distribution.
b. Now for 95% confidence level,
α = 0.05, α/2 = 0.025
From the t tables, T.INV.2T(α, degree of freedom), we find the t value as
t =T.INV.2T(0.05, 128) = 2.34
Taking the positive value of t, we get
Confidence interval is ,
(32.52, 35.8)
95% confidence interval is (32.52, 35.8)
So with 
confidence of the population of the mean number of the pounds per person per week is between 32.52 pounds and 35.8 pounds.
c. About of confidence intervals which contains the true population of mean number of the pounds of the trash that is generated per person per week and about 
that doe not contain the true population of mean number of the pounds of trashes generated by per person per week.
Answer:
250
Step-by-step explanation:
Pretty much every time it asks for the abasloute value it just asks how much to get the zero on the number line. It will ALWAYS be positive. For example -5 will be 5 to get to zero on the numberline. Same thing with a positive 5.
