Acertain type of stainless steel powder is supposed to have a mean particle diameter of µ=15 µm. a random sample of 87 particles had a mean diameter of 15.2 µm, with a standard deviation of 1.8 µm. a test is made of h0: μ=15 versus h1: μ≠15.

The z-score corresponding to the actual mean of x = 15.2 can be calculated using the formula: z = (x - mean) / (SD / sqrt(n) ), where n is the sample size.
z = (15.2 - 15) / (1.8/sqrt(87)) = 1.04
Based on a z-table, the probability that z > 1.04 (and thus, the probability that x > 15.2) is 0.1492. This corresponds to a significance value of 1 - 0.1492 = 0.8508. Therefore, if alpha > 0.8508 (for example, 90%, 95%, or 99% confidence interval), then we fail to reject the null hypothesis, and assume that there is no significant difference from the mean.

Neither is the answer