Questions: The average math SAT score is 514 with a standard deviation of 111. A particular high school claims that its students have unusually high math SAT scores. A random sample of 45 students from this school was selected, and the mean math SAT score was 561. Is the high school justified in its claim? Explain. Yes, because the z-score () is since it within the range of a usual event, namely within mean of the sample means.

The average math SAT score is 514 with a standard deviation of 111. A particular high school claims that its students have unusually high math SAT scores. A random sample of 45 students from this school was selected, and the mean math SAT score was 561. Is the high school justified in its claim? Explain.

Yes, because the z-score () is  since it  within the range of a usual event, namely within  mean of the sample means.
Transcript text: The average math SAT score is 514 with a standard deviation of 111. A particular high school claims that its students have unusually high math SAT scores. A random sample of 45 students from this school was selected, and the mean math SAT score was 561. Is the high school justified in its claim? Explain. Yes, because the z-score ( $\square$ ) is $\square$ since it $\square$ within the range of a usual event, namely within $\square$ mean of the sample means.
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Solution

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Solution Steps

Step 1: Calculate the Z-score

To determine if the high school students have unusually high math SAT scores, we first calculate the Z-score using the formula:

\[ z = \frac{X - \mu}{\sigma} \]

where:

  • \(X = 561\) (sample mean),
  • \(\mu = 514\) (population mean),
  • \(\sigma = \frac{111}{\sqrt{45}} \approx 16.5469\) (standard error).

Calculating the Z-score:

\[ z = \frac{561 - 514}{16.5469} \approx 2.84 \]

Step 2: Calculate the Standard Error

The standard error \(SE\) is calculated as follows:

\[ SE = \frac{\sigma}{\sqrt{n}} = \frac{111}{\sqrt{45}} \approx 16.55 \]

Step 3: Perform the Z-test

Next, we calculate the Z-test statistic:

\[ Z_{\text{test}} = \frac{\bar{x} - \mu_0}{SE} = \frac{561 - 514}{16.55} \approx 2.84 \]

For a right-tailed test, we find the p-value:

\[ P = 1 - T(z) \approx 0.0 \]

Step 4: Conclusion

Since the p-value \(0.0\) is less than the significance level \(\alpha = 0.05\), we reject the null hypothesis. Thus, we conclude:

Yes, because the Z-score is not within the range of a usual event, namely within 95% mean of the sample means.

Final Answer

\(\boxed{\text{Yes}}\)

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