t and p-value please. WILL thumbs up if correct. Consider the accompanying data on breaking load (kg/25 mm width) for various fabrics in both an unabraded condition and an abraded condition. Use the paired t-test to test \( H_0 : \mu_D = 0 \) versus \( H_a : \mu_D > 0 \) at a significance level of 0.01. (Use \( \mu_D = \mu_U - \mu_A \).)| | Fabric ||-----------|------------------|| | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 || U | 36.2 | 55.0 | 51.2 | 38.6 | 43.2 | 48.8 | 25.6 | 49.6 || A | 28.5 | 20.0 | 46.0 | 34.0 | 36.5 | 52.5 | 26.5 | 46.5 |Calculate the test statistic and determine the \( P \)-value. (Round your test statistic to one decimal place and your \( P \)-value to three decimal places.)\( t = \) [Input Box]\( P \)-value = [Input Box]Explanation: This table presents the breaking load data for eight different fabrics, comparing results in unabraded (U) and abraded (A) conditions. The task is to apply a paired t-test to see if there is a statistically significant difference in the breaking load between the two conditions, with the hypothesis \( H_0 \) indicating no difference and \( H_a \) suggesting a greater breaking load in the unabraded condition at a 1% significance level. The calculations require rounding the t-test result to one decimal place and the \( P \)-value to three decimal places.

It is required to obtain the test statistic and P-value using the paired t-test to test the given hypotheses.The quantity μD is the true mean difference, such that .The data are paired, as each fabric is subjected to both conditions.The null and alternative hypotheses are: versus .The alternative hypothesis indicates the test is right-tailed.Here, the sample size is, pairs, so that the degrees of freedom is, .The necessary calculations are shown below:FabricUADifference, 136.228.57.70.23765625255.020.035772.1451563351.246.05.24.05015625438.634.04.66.82515625543.236.56.70.26265625648.852.5–3.7119.0826563725.626.5–0.965.81265625849.646.53.116.91265625Total––The mean difference is:The sample standard deviation of the differences is:The sample mean of the differences is, .The sample standard deviation of the differences is, .The test statistic is obtained as follows:Thus, the value of the test statistic is 1.7.The P-value for the right-tailed test is:The Excel formula is:Thus, .Decision rule: If the P-value is less than or equal to the significance level α, reject H0. Otherwise, fail to reject H0.Here, the significance level is, .Since the P-value of 0.066 is greater than the significance level or α value of 0.01, fail to reject H0. There is not enough evidence to reject the null hypothesis, , that is, there is not enough evidence to suggest that the mean breaking load of fabrics is higher in an unabraded condition than in an abraded condition.The t-test statistic is 1.7.The P-value is 0.066.There is not enough evidence to conclude that the mean breaking load of fabrics is higher in an unabraded condition than in an abraded condition, at the significance level of 0.01.