To determine if there is a difference between the two sets of measurements, we perform a matched pairs t-test. The value of the standardized test statistic is -2.262 and falls within the rejection region. The p-value is less than 0.045, providing strong evidence against the null hypothesis.
To determine if there is a difference between the two sets of measurements, we can perform a matched pairs t-test. Let's set up the null and alternative hypotheses:
Null hypothesis (H0): There is no difference between the two sets of measurements (μd = 0)
Alternative hypothesis (Ha): There is a difference between the two sets of measurements (μd ≠ 0)
To calculate the test statistic, we use the formula: t = (mean of difference - hypothesized mean) / (standard deviation of difference / sqrt(n)), where n is the sample size.
In this case, the mean of difference is -1.2, the hypothesized mean is 0, and the standard deviation of difference is 7.6. The sample size is 58. Plugging in these values, we get t = (-1.2 - 0) / (7.6 / sqrt(58)). Calculating this, we find that the value of the standardized test statistic is -2.262.
To determine the rejection region, we need to find the critical value for a two-tailed test with alpha = 0.045 and 57 degrees of freedom (because n-1 = 58-1 = 57). Looking up the critical value in the t-table, we find that it is approximately ±2.001.
The p-value is the probability of observing a test statistic as extreme as the one calculated (or more extreme), assuming the null hypothesis is true. To find the p-value, we can use a t-table or a t-distribution calculator. For a two-tailed test, we multiply the obtained p-value by 2. In this case, the p-value is less than 0.045, indicating strong evidence against the null hypothesis.
Based on the calculated test statistic, the rejection region, and the p-value, we can conclude that there is evidence, at an alpha = 0.045 level of significance, to conclude that there is a difference between the two sets of measurements.