8 To compare the average swimming times for two swimmers, each swimmer was asked to swim freestyle for a distance of 100 yards at randomly selected times. The swimmers were thoroughly rested between laps and did not race against each other, so each sample of times was an independent random sample. The times for each of 10 trials are shown for the two swimmers.**Time Table:**| | Swimmer 1 | Swimmer 2 ||---------|-----------|-----------|| **1** | 59.61 | 59.75 || **2** | 59.93 | 59.42 || **3** | 59.84 | 59.54 || **4** | 59.49 | 59.62 || **5** | 59.56 | 59.49 || **6** | 59.45 | 59.62 || **7** | 59.71 | 59.67 || **8** | 60.06 | 59.39 || **9** | 59.64 | 59.65 || **10** | 59.63 | 59.50 |

Since you have posted a question with multiple subparts we will solve first three subparts for you. To get remaining subparts solved, please repost the question and mention the subparts to be solved. Hypothesis tests are conducted when larger populations aren't able to compare easily; in such cases, the samples are taken from the population and analyzed to check whether the sample resembles the population or not.Given: The swimming speed of two swimmers is given as: Swimmer 1 Swimmer 2 59.61 59.8 59.49 59.33 59.64 59.75 59.49 59.63 60 59.85 59.75 59.42 59.44 59.62 59.71 59.49 59.62 59.82 59.69 59.5 Claim: The average speed of swimmer 2 is faster than the average speed of swimmer 1. Let the null and alternate hypotheses be: H0:μ1-μ2=0H1:μ1-μ2<0 Hence, option A is correct. Incorrect options: Option B - The alternate hypothesis states that swimmer 2 swims less than swimmer 1. Option C - The alternate hypothesis is given as μ1-μ2>0. Option D - The null hypothesis has inequality. Option E - The alternate hypothesis has no difference.b) Using the given data: Swimmer 1 (x1) Swimmer 2 (x2) x12 x22 59.61 59.8 3553.352 3576.04 59.49 59.33 3539.06 3520.049 59.64 59.75 3556.93 3570.063 59.49 59.63 3539.06 3555.737 60 59.85 3600 3582.023 59.75 59.42 3570.063 3530.736 59.44 59.62 3533.114 3554.544 59.71 59.49 3565.284 3539.06 59.62 59.82 3554.544 3578.432 59.69 59.5 3562.896 3540.25 ∑x1=596.44 ∑x2=596.21 ∑x12=35574.3 ∑x22=35546.93 Using the table calculated values, the mean and standard deviation is calculated as : Mean: x¯1=∑x1n=596.4410=59.644 x¯2=∑x2n=596.2110=59.621 Variance: s12=∑x12n-∑x1n2=35574.310-596.44102=0.0233 s22=∑x22n-∑x2n2=35546.9310-596.21102=0.0294 Test statistic: t=x¯1-x¯2s12+s22n-1=59.644-59.6210.0233+0.029410-1=0.301 The test statistic is 0.301.c) Degree of freedom: df=n1+n2-2=10+10-2=18 The p value at 18 degrees of freedom for the test statistic 0.301 at 0.05 level of significance is 0.3834. Hence, option F is correct, i.e., p-value>0.100. Incorrect options: Option A - The p value is not less than 0.0008. Option B - The p value does not lie between 0.005 and 0.010. Option C - The p value does not lie between 0.010 and 0.025. Option D - The p value does not lie between 0.025 and 0.050. Option E - The p value does not lie between 0.90 and 0.100.