Statistic assignment !!!

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 48150.9 6 434.69 r = n ∑ xy − ∑ x ∑ y √ ( n ∑ x 2 − ( ∑ x ) 2 )( n ∑ y 2 − ( ∑ y ) 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 attempt s (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 1066022 5 565 4018 2270170 319225 1614432 4 528 3669 1937232 278784 1346156

conclude that there is a significant linear relationship between the data? H o : ρ = 0 H a : ρ≠ 0 Test statistics: 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. b = n ∑ xy − ∑ x ∑ y ¿¿ b = ( 7 ∗ 10574482 ) −( 2978 ∗ 21369 ) ( 7 ∗ 1458044 ) − ( 2978 ) 2 b = ¿ 7.762 a ¿ y − b x 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

b.  = 0.01 Given that p-value=0.0321and  =0.01, since 0.0321>0.01, we fail to reject the null hypothesis. 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= x − 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= x − 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

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.