Answer:
The P-value for this test is P=0.2415.
Step-by-step explanation:
We have to perform an hypothesis testing on the mean of alla account balances.
The claim is that the mean of all account balances is significantly greater than $1,150.
Then, the null and alternative hypothesis are:
The sample size is n=20, with a sample mean is 110 and standard deviation is 125.
We can calculate the t-statistic as:
The degrees of freedom fot this test are:
For this one-tailed test and 19 degrees of freedom, the P-value is:
Answer:
The answer is below
Step-by-step explanation:
The z score is used to determine by how many standard deviations the raw score is above or below the mean. The z score is given as:
Given that μ = $72,641 and σ = $30,000.
a) x > $100000
P(x > 100000) = P(z > 0.91) = 1 - 0.8186 = 0.1814
b) n = 5
x > $100000
P(x > 100000) = P(z > 2.04) = 1 - 0.9793 = 0.0207
c) n = 10
x > $100000
P(x > 100000) = P(z > 2.88) = 1 - 0.9980 = 0.002
150 divided by 25 is 6, so I think it would be 6 hours to equal the total cost.
Answer:
The answer for this question is C
