1. A Texas House representative proposed placing solar panels on all public schools to increase green energy and reduce energy costs for districts. Suppose a manufacturer’s panels will be purchased only if real world results of the proposed panels have an average efficiency rating greater than 18%. A random selection of 32 panels results in a mean efficiency rating of 0.182 with a standard deviation of 0.006. Is there sufficient evidence at the 5% significance level to purchase the proposed panels?

Yes, at 5% significant level, there is sufficient statistical evidence to purchase the panels

Step-by-step explanation:

The given parameters are;

The acceptable efficiency, μ > 18% = 0.18

The number of panels in the sample, n = 32 panels

The mean efficiency of the panels, = 0.182

The standard deviation of the sample, s = 0.006

The significance level of the test, α = 5%

The null hypothesis; H₀ p ≤ 0.18%

Given that the population standard deviation is unknown, we use the t-test statistic which is given as follows;

Plugging in the values we get;

The degrees of freedom, df = n - 1 = 32 - 1 = 31 (a df value of 30 is used as an approximation to obtain the value from the t-table)

The critical t at 5% significance level t = 1.697

Therefore, given that the test statistic is larger than the critical t, we reject the null hypothesis, and there is sufficient statistical evidence that efficiency of the panels is larger than 18% and therefore, we should purchase the panels

answer: 4 meters

step-by-step explanation: to you understand 2/3 better you can do 1 divided by 3 = . then multiply that by the 2 which is .66666666666666667 then you would do 6 x .66666666666666667 which is 4.      

hope this .