Answer:
s >= 11
Step-by-step explanation:
s-12 >= -1
Add 12 to both sides
s >= 11
Answer:
The 90% confidence interval for the mean test score is between 77.29 and 85.71.
Step-by-step explanation:
We have the standard deviation for the sample, so we use the t-distribution to solve this question.
The first step to solve this problem is finding how many degrees of freedom, we have. This is the sample size subtracted by 1. So
df = 25 - 1 = 24
90% confidence interval
Now, we have to find a value of T, which is found looking at the t table, with 24 degrees of freedom(y-axis) and a confidence level of
. So we have T = 2.064
The margin of error is:

In which s is the standard deviation of the sample and n is the size of the sample.
The lower end of the interval is the sample mean subtracted by M. So it is 81.5 - 4.21 = 77.29
The upper end of the interval is the sample mean added to M. So it is 81.5 + 4.21 = 85.71.
The 90% confidence interval for the mean test score is between 77.29 and 85.71.
Answer:
Sam has the more convincing victory with a greater Zscore value
Step-by-step explanation:
Given that :
Year 1:
Mean finish time (m) = 185.64
Standard deviation (s) = 0.314
Sam's time (x1) = 185.29
Year 2:
Mean finish time (m) = 110.3
Standard deviation (s) = 0.129
Rita's time (x1) = 110.02
Zscore = (x - mean) / standard deviation
Sam's Zscore :
(185.29 - 185.64) / 0.314
= - 0.35 / 0.314
= −1.114649
= - 1.115
Rita's Zscore :
(110.02 - 110.3) / 0.129
= - 0.28 / 0.129
= −2.170542
= - 2.171
Sam has the more convincing victory with a greater Zscore value