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 Station 1 2 3 4 5 January 135 124 128 64 78 April 108 111 100 88 61 Does this information indicate that the peak wind gusts are higher in January than in April? Use ? = 0.01. (Let d = January − April.) (a) What is the level of significance? State the null and alternate hypotheses. Will you use a left-tailed, right-tailed, or two-tailed test? H0: ?d = 0; H1: ?d < 0; left-tailedH0: ?d > 0; H1: ?d = 0; right-tailed    H0: ?d = 0; H1: ?d > 0; right-tailedH0: ?d = 0; H1: ?d ≠ 0; two-tailed (b) What sampling distribution will you use? What assumptions are you making? The standard normal. We assume that d has an approximately normal distribution.The standard normal. We assume that d has an approximately uniform distribution.    The Student's t. We assume that d has an approximately normal distribution.The Student's t. 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. P-value > 0.2500.125 < P-value < 0.250    0.050 < P-value < 0.1250.025 < P-value < 0.0500.005 < P-value < 0.025P-value < 0.005 Sketch the sampling distribution and show the area corresponding to the P-value

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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 Station 1 2 3 4 5
January 135 124 128 64 78
April 108 111 100 88 61

Does this information indicate that the peak wind gusts are higher in January than in April? Use ? = 0.01. (Let d = January − April.)

(a) What is the level of significance?


State the null and alternate hypotheses. Will you use a left-tailed, right-tailed, or two-tailed test?
H0: ?d = 0; H1: ?d < 0; left-tailedH0: ?d > 0; H1: ?d = 0; right-tailed    H0: ?d = 0; H1: ?d > 0; right-tailedH0: ?d = 0; H1: ?d ≠ 0; two-tailed

(b) What sampling distribution will you use? What assumptions are you making?
The standard normal. We assume that d has an approximately normal distribution.The standard normal. We assume that d has an approximately uniform distribution.    The Student's t. We assume that d has an approximately normal distribution.The Student's t. 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.
P-value > 0.2500.125 < P-value < 0.250    0.050 < P-value < 0.1250.025 < P-value < 0.0500.005 < P-value < 0.025P-value < 0.005

Sketch the sampling distribution and show the area corresponding to the P-value.
   
   

(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 ??
At the ? = 0.01 level, we fail to reject the null hypothesis and conclude the data are not statistically significant.At the ? = 0.01 level, we fail to reject the null hypothesis and conclude the data are statistically significant.    At the ? = 0.01 level, we reject the null hypothesis and conclude the data are not statistically significant.At the ? = 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.
Reject the null hypothesis, there is sufficient evidence to claim average peak wind gusts are higher in January.Fail to reject the null hypothesis, there is insufficient evidence to claim average peak wind gusts are higher in January.    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.
Expert Solution
Step 1

Note: " Since you have posted many sub-parts. we will solve the first three sub-parts for you. To get the remaining sub-parts solved, please resubmit the complete question and mention the sub-parts to be solved."

Given: Five weather stations on Trail Ridge Road in Rocky Mountain National Park, the peak wind gusts (in miles per hour) for January and April.

Let x denote the peak wind gusts for January.

Let y denote the peak wind gusts for April.

Let d denote the difference of peak wind gusts for January and April (di=xi-yi).

Let μd denotes the difference of mean for the peak wind gusts for January and April.

The statistical hypothesis formed will be:

H0: μd=0

H1: μd>0

The test is right tailed test.

(a) The level of significance(α) is 0.01. 

Step 2

(b)  The sampling distribution that will be used is "The Student's t".

In this scenario, the assumptions needed to satisfy before conducting the paired t test are:

1) Samples for the peak wind gusts for January and April are not independent.

2) The population of differences follows normal distribution.

3) Population variances are unknown but not necessarily equal.

4) Size of both samples are equal.

So, we assume that d has an approximately normal distribution.

Therefore, the correct option is that sampling distribution is the Student's t and we assume that d has an approximately normal distribution.

The value of test statistics is given as:

                                                 t=d¯sdn  =12.221.22975  =1.284where,d¯=i=15(xi-yi)n  =(135-108)+(124-111)+...+(78-61)5  =12.2sd=i=15(di-d¯)2n-1   =((135-108)-12.2)2+((124-111)-12.2)2+...+((78-61)-12.2)25-1   =21.2297

Hence, the value of sample test statistics is 1.284.

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