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In national vocabulary test, it is known that the mean score is 68. A class of 20 students takes the test and has a mean score of 64 and the standard deviation is 13. Is there evidence that the average of this class is significantly different from the national average?

Use the 4 step process. (State, Formulate, Solve, Conclude)

User Sksallaj
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Final answer:

Based on the calculated test statistic and the critical value, we can make a conclusion about whether there is evidence that the average score of the class is significantly different from the national average.

Step-by-step explanation:

To determine if there is evidence that the average score of a class is significantly different from the national average on a vocabulary test, we can follow a four-step hypothesis testing process:

Step 1): State: The null hypothesis (νo) states that there is no significant difference between the class mean and the national mean, so νo: μ = 68.

The alternative hypothesis (α) posits that there is a significant difference, so α: μ ≠ 68.

Step 2): Formulate: We will use a two-tailed t-test to test the hypothesis since the population standard deviation is unknown and the sample size is less than 30.

The t-score is calculated using the formula (μx - μ) / (s/ √n), where μx is the sample mean, μ is the population mean, s is the sample standard deviation, and n is the sample size.

Step 3): Solve: For this class: μx = 64, s = 13, n = 20, and μ = 68. Substituting these values into the formula gives us a t-score.

We then consult a t-distribution table to find the p-value associated with this t-score at a 95% confidence level.

Step 4): Conclude: If the p-value is less than the significance level (typically 0.05), we reject the null hypothesis, concluding that there is a significant difference in scores.

Otherwise, we fail to reject the null hypothesis, indicating that there is no significant evidence of a difference in scores.

User YWE
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