t and p-value please. WILL thumbs up if correct. Torsion during hip external rotation (ER) and extension may be responsible for certain kinds of injuries in golfers and other athletes. The article "Hip Rotational Velocities During the Full Golf Swing reported on a study in which peak ER velocity and peak IR (internal rotation) velocity (both in deg-sec-¹) were determined for a sample of 15 female collegiate golfers during their swings. The following data was supby the article's authors.Golfer ER123456789101112131415diff z percIR-130.6 -98.9 -31.7 -1.28-125.1-115.9 -9.2 -0.97-51.7 -161.6 109.9 0.34-179.7 -196.9 17.2 -0.73-130.5 -170.7 40.2 -0.34-101.0 -274.9 173.9 0.97-24.4 -275.0 250.6 1.83-231.1 -275.7 44.6 -0.17-186.8 -214.6 27.8 -0.52-58.5 -117.8 59.3 0.00-219.3 -326.7 107.4 0.17LAUSE SALT-113.1 -272.9 159.8 0.73-244.3 -429.1 184.8 1.28-184.4 -140.6 -43.8 -1.83-199.2 -345.6 146.4146.4 0.52(a) Is it plausible that the differences came from a normally distributed population?Yes, it's plausible that the population distribution of differences is normal.O No, it's not plausible that the population distribution of differences is normal.(b) The article reported that mean (SD) -145.3(68.0) for ER velocity and -227.8(96.6) for IR velocity. Based just on this information, could a test of hypotheses about the difference between true average IR velocity and true average ER velocity be carried out? Explain.O Yes, since the data is not paired, the sample means and standard deviations are useful summaries for inferenceYes, since the data is paired, the sample means and standard deviations are useful summaries for inference.H₂ HD 0ⒸH₂i Ho-OO No, since the data is not paired, the sample means and standard deviations are not useful summaries for inference.No, since the data is paired, the sample means and standard deviations are not useful summaries for inference.OH₂H₂O(c) The article stated that "The lead hip peak IR velocity was significantly greater than the trail hip ER velocity (p = 0.003, t value - 3.65)." (The phrasing suggests that an upper-tailed test was used.) Is that in fact the case? [Note: "p=0.033" in Table 2 of the article is erroneous.]State and test the appropriate hypotheses. (Use a = 0.05.)OH₂M₂=OH: MosoⒸM₂² M₂-0Calculate the test statistic value and P-value. (Round your test statistic to one decimal place and your P-value to three decimal places.)P-value=

GolferERIRdiff1-130.6-98.9-31.72-125.1-115.9-9.23-51.7-161.6109.94-179.7-196.917.25-130.5-170.740.26-101-274.9173.97-24.4-275250.68-231.1-275.744.69-186.8-214.627.810-58.5-117.859.311-219.3-326.7107.412-113.1-272.9159.813-244.3-429.1184.814-184.4-140.6-43.815-199.2-345.6146.4The relevant null and alternative hypotheses are: i.e., average difference between ER velocity and IR velocity is zero. i.e., average difference between ER velocity and IR velocity is greater than zero.where, Sample mean of the differences is computed as follows:Sample standard deviation of the differences is computed as follows:Substitute the values in the formula:Thus, the value of the test statistic is 3.7. (rounded to one decimal place).P-value is the probability of obtaining the test statistic value or a value even more extreme, assuming the null hypothesis is true.Degrees of freedom:Use Excel function "=T.DIST.RT(3.655,14)" to obtain the P-value for right-tailed test. The result is shown below:Thus, the P-value is 0.001. (rounded to three decimal places)Conclusion:Since the P-value is less than the significance level 0.05, reject the null hypothesis. Therefore, there is enough evidence to conclude that average difference between ER velocity and IR velocity is greater than zero.The test statistic is P-value is 0.001.