Final answer:
We use a one-sample t-test to determine if the mean price of 7.5-cubic-foot refrigerators is greater than $300. The t-statistic is calculated and compared to critical t-values or a p-value is computed. If the p-value is less than the level of significance, we reject the null hypothesis.
Step-by-step explanation:
To evaluate whether the data provide convincing evidence that the mean price of 7.5-cubic-foot refrigerators is greater than $300, we can use a one-sample t-test. Given the sample mean ($310.9), sample variance (966.3222), and the sample size (10), we first state the null hypothesis (H₀: μ ≤ $300) and the alternative hypothesis (H₁: μ > $300).
At common levels of significance (such as 0.05), we can calculate the t-statistic using the sample mean, the hypothesized mean, and the sample standard deviation (the square root of the variance). The t-statistic is computed as (Sample mean - Hypothesized mean) / (Sample standard deviation / sqrt(n)).
Using this formula, we find that the t-statistic is ($310.9 - $300) / (sqrt(966.3222) / sqrt(10)), which is a positive number. We would then compare this t-statistic to critical t-values from the t-distribution table or compute the p-value. If the p-value is less than the chosen level of significance, we reject the null hypothesis and conclude that there is sufficient evidence that the mean price is greater than $300. If the p-value is greater, we do not reject the null hypothesis.