0 of 1 Point
math teacher claims that she has developed a review course that increases the scores of students on the math portion of a college entrance exam. Base in data from the administrator of the exam, scores are normally distributed with $\mu=515$. The teacher obtains a random sample of 2200 students, puts then hrough the review class, and finds that the mean math score of the 2200 students is 522 with a standard deviation of 116 . Complete parts (a) through (d) below.
\[
t_{0}=2.83
\]
(Round to two decimal places as needed.)
Find the P-value.
The $P$-value is $\square$.
(Round to three decimal places as needed.)
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Answer
Rounding to three decimal places, the final P-value is \(\boxed{0.005}\).
Step 1 ：The degrees of freedom for the t-distribution is the sample size minus 1, which is 2200 - 1 = 2199.
Step 2 ：We can use this value along with the t-value to calculate the P-value.
Step 3 ：The t-value is 2.83 and the degrees of freedom is 2199.
Step 4 ：The P-value is calculated as 0.004697063306862592.
Step 5 ：Rounding to three decimal places, the final P-value is \(\boxed{0.005}\).
