Correlation Analysis of Wildland Fires and Passing Yards

1. The number of Wildland fires (in thousands) and wildland acres burned (in millions) in the US for eight years is in the table below, Fires (x) 84.1 73.5 63.6 65.5 66.8 96.4 85.7 79.0 Acres (y) 3.6 7.2 4.0 8.1 8.7 9.9 9.3 5.3 a. Calculate the sample correlation coefficient, r. Round to the nearest 1000 th . Fires (X) Acres (Y) XY X 2 Y 2 84.1 3.6 302.76 7072.81 12.96 73.5 7.2 529.2 5402.25 51.84 63.6 4 254.4 4044.96 16 65.5 8.1 530.55 4290.25 65.61 66.8 8.7 581.16 4462.24 75.69 96.4 9.9 954.36 9292.96 98.01 85.7 9.3 797.01 7344.49 86.49 79 5.3 418.7 6241 28.09 614.6 56.1 4368.14 48150.96 434.69 𝑟 = 𝑛 ∑ ?? − ∑ ? ∑ ? √(𝑛 ∑ ? 2 − (∑ ?) 2 )(𝑛 ∑ ? 2 − (∑ ?) 2 ) 𝑟 = (8 ∗ 4368.14) − (614.6 ∗ 56.1) √((8 ∗ 48150.96) − (614.6) 2 )((8 ∗ 434.69) − (56.1) 2 ) 𝑟 = 0.297

b. Describe the type of correlation coefficient and interpret the correlation in the context of the data. The correlation coefficient ( r ) of 0.297 reveals a positive correlation between the number of wildland fires and the wildland acres burned in the context of the provided data. This implies that, on average, as the number of wildland fires increases, there is a tendency for a rise in the wildland acres burned. 2. The numbers of pass attempts and passing yards for seven professional quarterback for a recent year are listed in the table below. Round all answers to the nearest 1000 th . Pass attempts (x) 449 565 528 197 670 351 218 Passing Yards (y) 3265 4018 3669 1141 5177 2362 1737 a. Calculate the sample correlation coefficient, r. Pass attempts (X) Passing Yards (Y) XY X 2 Y 2 449 3265 1465985 201601 10660225 565 4018 2270170 319225 16144324 528 3669 1937232 278784 13461561 197 1141 224777 38809 1301881 670 5177 3468590 448900 26801329 351 2362 829062 123201 5579044 218 1737 378666 47524 3017169

2978 21369 10574482 1458044 76965533 𝑟 = 𝑛 ∑ ?? − ∑ ? ∑ ? √(𝑛 ∑ ? 2 − (∑ ?) 2 )(𝑛 ∑ ? 2 − (∑ ?) 2 ) 𝑟 = (7 ∗ 10574482) − (2978 ∗ 21369) √((7 ∗ 1458044) − (2978) 2 )((7 ∗ 76965533) − (21369) 2 ) 𝑟 = 0.991 b. Describe the type of correlation coefficient and interpret the correlation in the context of the data. The correlation coefficient of 0.991 between pass attempts and passing yards for the seven professional quarterbacks indicates an exceptionally strong positive correlation. This implies that as the number of pass attempts increases, there is a robust tendency for passing yards to also increase. c. Test the significance of the correlation coefficient, r, that you found in part a. Use α=.05. Is there enough evidence to conclude that there is a significant linear relationship between the data? 𝐻 𝑜 : 𝜌 = 0 𝐻 𝑎 : 𝜌 ≠ 0 Test statistics:

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𝑡 = 𝑟 ∗ √ 𝑛 − 2 1 − 𝑟 2 𝑡 = 0.991 ∗ √ 7 − 2 1 − 0.991 2 𝑡 = 16.28 Degrees of freedom =n-2= 7-2 =5 The P-value for 2-tailed test using Excel =T.DIST.2T(16.28,5)=0 Since the P-value (0) is less than alpha(0.05) we reject the null hypothesis. Since the null hypothesis is rejected, the correlation is significant d. Find the equation of the regression line for the data. ? = 𝑛 ∑ ?? − ∑ ? ∑ ? (𝑛 ∑ ? 2 − (∑ ?) 2 ? = (7 ∗ 10574482) − (2978 ∗ 21369) (7 ∗ 1458044) − (2978) 2 ? = 7.762 a = ? ̅ − ??̅ ? = ( 21369 7 ) − (7.762 ∗ 2978 7 ) a= -249.46 Equation of the regression: y= 7.762x – 249.46 e. Use the regression equation to predict the average number passing yards if the pass attempts are 250.

y= 7.762x – 249.46 y= 7.762(250) – 249.46 y= 1691.04 3. Write each claim as a mathematical sentence. State the null and alternative hypotheses, identify which represents the claim, and state which whether the test would be a left-tailed test, a right-tailed test or a two-tailed test . a. A school publicizes that the proportion of its students who earn scholarships to College is 85% The null and alternative hypotheses are H 0 : 𝑝 = 0.85 H 1 : 𝑝 ≠ 0.85 This is a two-tailed test. H 0 is the claim. b. A grocery store says that the mean time wait in a line is less than 5 minutes.

The null and alternative hypotheses are H 0 : μ = 5 H 1 : μ < 5 This is a left-tailed test. H 1 is the claim. c. A company advertises that the mean life of its furnaces is at least 15 years. The null and alternative hypotheses are H 0 : μ < 15 H 1 : μ ≥ 5 This is a Right-tailed test. H 0 is the claim. 4. The P -value for a hypothesis test is P = 0.0321. What is your decision if the level of significance is: a.  = 0.05? Given that p-value=0.0321and  =0.05, since 0.0321≤0 .050, we reject the null hypothesis. b.  = 0.01 Given that p-value=0.0321and  =0.01, since 0.0321>0.01, we fail to reject the null hypothesis.

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5. In auto racing, a pit crew claims that its mean pit stop time (for 4 new tires and fuel) is less than 13 seconds. A random selection of 32 pit stop times has a sample mean of 12.9 seconds. Assume the population standard deviation is 0.19 second. Is there enough evidence to support the claim at  = 0.01? Use a P -value. The null hypothesis ( H 0) is that the mean pit stop time is 13 seconds The alternative hypothesis ( Ha ) is that the mean pit stop time is less than 13 seconds. Z= 𝑥̅−𝜇 𝜎 √𝑛 Z= 12.9−13 0.19 √32 Z= -2.98 p- value= 0.0014 p-value=0.0014 <  =0.01 Since the p-value is less than the significance level, we reject the null hypothesis ( H 0). There is enough evidence to support the claim that the mean pit stop time is less than 13 seconds at the 0.01 significance level. 6. An inventor has developed a new, energy efficient lawn mower engine. He claims that the engine will run continuously for 300 minutes on a single gallon of regular gasoline. From his stock of 2000 engines, the inventor selects a simple random sample of 50 engines for testing and finds the sample mean to be 295 minutes and sample standard deviation to be 20 minutes. Is there

enough evidence to support his claim at the 5% level of significance? Assume the mean times are normally distributed. H 0: The mean running time of the engine is 300 minutes ( μ =300). H 1: The mean running time of the engine is less than 300 minutes ( μ <300). t= 𝑥̅−𝜇 𝑠 √𝑛 t= 295−300 20 √50 t= -1.768 Next, you need to find the critical t-value for a one-tailed test with 49 degrees of freedom (50 - 1) and a significance level of 0.05. p- value is 0.007908 t = − 1.798 is in the lower tail of the distribution and the p -value (0.007908) is less than 0.05, we have enough evidence to reject the null hypothesis. Therefore, we can conclude that there is evidence to support the inventor's claim that the engine will run for less than 300 minutes on a single gallon of regular gasoline at the 5% level of significance. 7. A medical researcher says that less than 23% of US adults are smokers. In a random sample of 200 US adults, 19% say that they are smokers. At 𝛼 = .05 , is there enough evidence to support the researchers claim? 𝑝 = 0.23, 𝑝̂ = 0.19 , n=20, ∝= 0.05

Hypotheses 𝐻 𝑜 : 𝑝 = 0.23 𝐻 1 : 𝑝 < 0.23 Test statistic ? = 𝑝̂ − 𝑝 √ 𝑝(1 − 𝑃) 𝑛 = 0.19 − 0.23 √ 0.23(1 − 0.23) 200 = −1.344 Z= -1.344 P- value = 0.0895 The P-value (0.0895) is greater than the level of significance (0.05) Hence we fail to reject the null hypothesis and conclude that there is no significant evidence to support the claim that less than 23% of US adults are smokers.

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Statistic assignment !!!

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