Sketch the sampling distribution and show the area corresponding to the P-value.0220022(d) Based on your answers in parts (a) to (c), will you reject or fail to reject the null hypothesis? Are the data statistically significant at level a?At the α = 0.01 level, we fail to reject the null hypothesis and conclude the data are statistically significant.At the α = 0.01 level, we reject the null hypothesis and conclude the data are not statistically significant.At the α = 0.01 level, we fail to reject the null hypothesis and conclude the data are not statistically significant.At the α = 0.01 level, we reject the null hypothesis and conclude the data are statistically significant.(e) State your conclusion in the context of the application.O Fail to reject the null hypothesis, there is insufficient evidence to claim average peak wind gusts are higher in January.O Reject the null hypothesis, there is sufficient evidence to claim average peak wind gusts are higher in January.O Reject the null hypothesis, there is insufficient evidence to claim average peak wind gusts are higher in January.Fail to reject the null hypothesis, there is sufficient evidence to claim average peak wind gusts are higher in January. In this problem, assume that the distribution of differences is approximately normal. Note: For degrees of freedom d.f. not in the Student's t table, use the closest d.f. that is smaller. In some situations, this choice of d.f. may increase the P-value by a small amount and therefore produce a slightly more "conservative" answer.At five weather stations on Trail Ridge Road in Rocky Mountain National Park, the peak wind gusts (in miles per hour) for January and April are recorded below.Weather StationJanuaryAprilUSE SALTDoes this information indicate that the peak wind gusts are higher in January than in April? Use α = 0.01. (Let d = January - April.)(a) What is the level of significance?12345137 120 126 64 78104 115 102 88 61State the null and alternate hypotheses. Will you use a left-tailed, right-tailed, or two-tailed test?Ho: My = 0₁ H₁:0; two-tailedO Hoi Hd O1 H₁: MO Hot Hd = 0; H₁:O Hot Hd = 0; H₁²= 0; right-tailed< 0; left-tailed> 0; right-tailed(b) What sampling distribution will you use? What assumptions are you making?O The standard normal. We assume that d has an approximately normal distribution.O The Student's t. We assume that d has an approximately normal distribution.O The Student's t. We assume that d has an approximately uniform distribution.O The standard normal. We assume that d has an approximately uniform distribution.What is the value of the sample test statistic? (Round your answer to three decimal places.)(c) Find (or estimate) the P-value.OP-value > 0.250O 0.125 < P-value < 0.250O 0.050 < P-value < 0.125O 0.025 < P-value < 0.050O 0.005 < P-value < 0.025O P-value < 0.005

a. Suppose μd is the population mean difference of wind gust between January and April. That, is μd = μJanuary – μApril.It is given that, α = 0.01. Thus, the level of significance is 0.01. It is needed to check, whether the peak wind gusts are higher in January than April. The null and alternative hypotheses are: H0: μd = 0: H1: μd > 0; right tailed. b) The peak wind gusts are normally distributed. And the observations are selected randomly and independently. The observations of peak wind gusts are considered on two different months. The Student’s t sampling distribution will be used, we assume that that d has an approximately normal distribution. The sample size is n = 5. Consider the below table. The test statistic is : The value of sample test statistic is 1.114. c) The degrees of freedom for the test statistic is, df = n – 1 = 5 – 1 = 4. Hence, the P-value using the EXCEL formula =T.DIST.RT(1.114,4), is 0.1639. The excel output is, From t table corresponding to degrees of freedom of 4 and level of significance of 0.20 (one tail), it is obtained that t0.20 = 0.941 and from t table corresponding to degrees of freedom of 4 and level of significance of 0.15 (one tail), it is obtained that t0.15 = 1.190. The test statistic of 1.114 lies within 0.941 and 1.190. Thus, the smallest range of the P-value is 0.15 to 0.20. Thus, the range is 0.125 < P-value < 0.250. The graph is, d) Decision rule: If P-value is less than or equal to the level of significance, reject the null hypothesis. Otherwise fail to reject the null hypothesis. Conclusion: Here, the P -value of 0.1639 is greater that the level of significance of 0.01. Hence, fail to reject the null hypothesis. At the α = 0.01 level, we fail to reject the null hypothesis and conclude that the data are not statistically significant. e) Fail to reject the null hypothesis, there is insufficient evidence to claim average peak wind gusts are higher in January.