A 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. Based on data from the administrator of the exam, scores are normally distributed with $\mu=515$. The teacher obtains a random sample of 1800 students, puts them through the review class, and finds that the mean math score of the 1800 students is 521 with a standard deviation of 113 . Complete parts (a) through (d) below
(d) Test the hypothesis at the $\alpha=0.10$ level of significance with $n=375$ students. Assume that the sample mean is still 521 and the sample standard deviation is still 113. Is a sample mean of 521 significantly more than 515 ? Conduct a hypothesis test using the P-value approach.

Find the test statistic
\[
t_{0}=1.03
\]
(Round to two decimal places as needed.)
Find the P-value
The $P$-value is
(Round to three decimal places as needed.)

Step 1 ：We are given the sample mean (521), the sample standard deviation (113), the sample size (375), and the test statistic (1.03). The null hypothesis is that the mean score is 515, and the alternative hypothesis is that the mean score is more than 515.

Step 2 ：We are conducting a one-tailed test because we are testing if the mean score is significantly more than 515. The P-value is the probability of observing a test statistic as extreme as the one calculated, assuming the null hypothesis is true.

Step 3 ：The P-value can be found using the cumulative distribution function (CDF) of the t-distribution. The CDF gives the probability that a random variable drawn from the t-distribution is less than or equal to a given value. Since we are conducting a one-tailed test, we need to find the probability that the t-statistic is greater than or equal to the calculated test statistic. This is equal to 1 minus the CDF of the t-distribution at the test statistic.

Step 4 ：Using the given values, we find that the P-value is approximately 0.152.

Step 5 ：The P-value is greater than the significance level of 0.10. This means that we do not reject the null hypothesis. There is not enough evidence to support the claim that the mean score is significantly more than 515.

Step 6 ：Final Answer: The P-value is approximately \(\boxed{0.152}\).