Select the correct answer from each drop-down menu. Simplify the given polynomial and use it to complete the statement. The poly
1 answer:
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Answer:
Test statistic t = 5.25
P-value = 0.000002 (one-tailed test)
Step-by-step explanation:
This is a hypothesis test for the population mean.
The claim is that the average amount of time that a freshman at TAMU studies is significantly higher than 7 hours.
Then, the null and alternative hypothesis are:
The significance level is 0.05.
The sample has a size n=49.
The sample mean is M=8.5.
As the standard deviation of the population is not known, we estimate it with the sample standard deviation, that has a value of s=√s^2=√4=2.
The estimated standard error of the mean is computed using the formula:
Then, we can calculate the t-statistic as:
The degrees of freedom for this sample size are:
This test is a right-tailed test, with 48 degrees of freedom and t=5.25, so the P-value for this test is calculated as (using a t-table):
As the P-value (0.000002) is smaller than the significance level (0.05), the effect is significant.
The null hypothesis is rejected.
There is enough evidence to support the claim that the average amount of time that a freshman at TAMU studies is significantly higher than 7 hours.
Answer:
Step-by-step explanation:
Given
Required
Determine the hours of practice over 6 days
First, we calculate the hours for the given 3 days.
The hours of practice for the other three days is also 52 hours.
So, the total is: