Question

The average score for a sample of 45 of your students on a quiz was 52, while other sections that took the same quiz was 58. The standard deviation for your class’s scores is 3.7. Can you conclude that the population mean score for your class is different than the mean score for the rest of the sections?

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Answer to a math question The average score for a sample of 45 of your students on a quiz was 52, while other sections that took the same quiz was 58. The standard deviation for your class’s scores is 3.7. Can you conclude that the population mean score for your class is different than the mean score for the rest of the sections?

Birdie
4.5
To determine whether the population mean score for the class is different than the mean score for the other sections, we can perform a t-test.

Given:
Sample size, n = 45
Sample mean, \bar{x} = 52
Population mean $\mu$ = 58 $mean score for the rest of the sections$
Sample standard deviation, s = 3.7

First, we need to calculate the t-score:
t = \frac{$\bar{x} - \mu$}{$s/\sqrt{n}$}

t = \frac{$52 - 58$}{$3.7/\sqrt{45}$}

t = \frac{-6}{$3.7/\sqrt{45}$}

t = \frac{-6}{$3.7/6.7082$}

t = \frac{-6}{0.5524}

t\approx-10.8782

Next, we need to find the critical t-value for a 2-tailed test at a 5% significance level with degrees of freedom $df$ = n - 1 = 45 - 1 = 44. You can look up this value in a t-distribution table or use statistical software.

For a 2-tailed test at a 5% significance level with df = 44, the critical t-value is approximately \pm2.0167 .

Since the absolute value of the calculated t-score $|10.8782|$ is greater than the critical t-value $2.0167$, we reject the null hypothesis.

Therefore, we can conclude that the population mean score for the class is different than the mean score for the rest of the sections.

\boxed{\text{Answer: Yes, you can conclude that the population mean score for your class is different than the mean score for the rest of the sections.}}

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