Question

The national mean SAT score in math is 550. Suppose a high school principal claims that the mean SAT score in math at his school is better than the national mean score. A random sample of 72 students finds a mean score of 574. Assume that the population standard deviation is sigma=100. Is the principal's claim valid? Use a level of significance of alpha=0.05.
Compute the test statistic for this analysis. Round your answer to 3 decimal places.
Z=
Determine the P-value based the test statistic. Round your answer to 3 decimal places.
P-value=
State your decision based on the P-value and the level of significance (alpha) and give your conclusion in an English sentence.
-Reject the null hypothesis. There is insufficient evidence to suggest that the students' mean SAT score is greater than 550. The principal was right.
-Reject the null hypothesis. There is sufficient evidence to suggest that the students' mean SAT score is greater than 550. The principal was right.
-Fail to reject the null hypothesis. There is insufficient evidence to suggest that the students' mean SAT score is greater than 550. The principal was wrong.
-Fail to reject the null hypothesis. There is sufficient evidence to suggest that the students' mean SAT score is greater than 550. The principal was wrong.

Answer 1

Answer:

Option B) Reject the null hypothesis. There is sufficient evidence to suggest that the students' mean SAT score is greater than 550. The principal was right.

Test statistics = 2.036

P- value = 2.07% .

Step-by-step explanation:

We are given that the population mean, $\mu$ = 550  {National mean SAT score}

We have to test the principal claim that the mean SAT score in math at his school is better than the national mean score or not.

Let              Null Hypothesis, $H_0$ : $\mu$ = 550

          Alternate Hypothesis, $H_1$ : $\mu$ > 550

The test statistics we will use here is;

             $\frac{Xbar - \mu}{\frac{\sigma}{\sqrt{n} } }$ follows N(0,1)

     where, Xbar = 574    and $\sigma$(Population standard deviation) = 100

                  n = sample of students = 72

 Test Statistics =  $\frac{574 - 550}{\frac{100}{\sqrt{72} } }$ = 2.036

Now at 5% level of significance the z table gives the critical value of 1.6449 and our test statistics is more than this as 2.036 > 1.6449 so we have sufficient evidence to reject null hypothesis and conclude that there is sufficient evidence to suggest that the students' mean SAT score is greater than 550. The principal was right.

P - value = P(Z > 2.04) = 0.0207 or 2.07%

Here also our P- value is less than the level of significance of 5% ,we will reject null hypothesis.