Question

An instructor whishes to see whether the variation in scores of the 23 students in her class is less than the variance of the population. The variance of the class is 198. Is there enough evidence to support the claim that the variation of the students is less than the population variance (�� 2 = 225) at α = 0.05?. Assume that the scores are normally distributed

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1. State the null and alternative hypotheses:

H_0: \sigma^2 = 225

H_a: \sigma^2 < 225

2. Identify sample size, sample variance, population variance, and significance level:

n = 23

s^2 = 198

\sigma_0^2 = 225

\alpha = 0.05

3. Calculate the Chi-Square test statistic:

\chi^2 = \frac{(23-1)s^2}{\sigma_0^2} = \frac{22 \cdot 198}{225} \approx 19.36

4. Determine the critical value from the Chi-Square distribution table:

\chi^2_{0.05, 22} \approx 12.338

5. Compare the test statistic to the critical value:

19.36 > 12.338

6. Make a decision:

\text{Do not reject } H_0

Answer:

\text{There is not enough evidence to support the claim that the variation of the students is less than the population variance at } \alpha = 0.05

2. Identify sample size, sample variance, population variance, and significance level:

3. Calculate the Chi-Square test statistic:

4. Determine the critical value from the Chi-Square distribution table:

5. Compare the test statistic to the critical value:

6. Make a decision:

Answer:

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