A taxi company has been using Brand A tires, and the distribution of kilometers to wear-out has been found to be approximately normal with μ = 114,000 and σ= 11,600. Now it tries 12 tires of Brand B...

A taxi company has been using Brand A tires, and the distribution of kilometers to wear-out has been found to be approximately normal with μ = 114,000 and σ= 11,600. Now it tries 12 tires of Brand B and finds a sample mean ofx= 117,200. Test at the 5% level of significance to see whether there is a significant difference (positive or negative) in kilometers to wear-out between Brand A and Brand B. Assume the standard deviation is unchanged. Show all steps of the procedure described before see below Example

Example: It is very important that a certain solution in a chemical process have a pH of 8.30. The method used gives measurements which are approximately normally distributed about the actual pH of the solution with a known standard deviation of 0.020. We decide to use 5% as the critical level of significance.

a) Suppose a single determination shows pH of 8.32. The null hypothesis is that the true pH is 8.30 (H0: pH = 8.30). The alternative hypothesis isHa: pH ≠8.30 (this is a two-sided test because there is no indication that changes in only one direction are important).

b) Suppose that now our sample consists of 4 determinations giving values of 8.31, 8.34, 8.32, 8.31. The sample mean isx
₁ = 8.32.