Question

Can someone help me with these? I have the answer's below, but I don't know how to solve them (theres a cumulative standard normal distribution table we should refer to for some of these, so if you do, could you tell me where in the table to look to0?) Thank you so much! d) Two catalysts were tested for their effect on a chemical reaction. Six different measurements were performed on the reaction using each catalyst. The conversion at 1 hour using catalyst A had a mean of 90% and a standard deviation of 3%; for catalyst B, the mean was 95% and a standard deviation of 7%. Determine whether the catalyst made a significant difference on the conversion at 1 hours within 90% confidence interval. A spectrometer is used on two successive days to measure the carbon content of a silicon wafer in ppm. On day 1, the measured values were 2.2634, 2.1904, 2.2643, 2.10204 2.2935. On day 2, the measured values were 2.4793, 2.1939, 2.2541, 2.3099. Can you conclude that the calibration of the spectrometer has changed from day 1 to day 2? Assume that while the calibrating point of the spectrometer may have changed, the variance of the spectrometer has not. Use a significance of 0.05. e)

Answer 1

Solution:-

d)

State the hypotheses. The first step is to state the null hypothesis and an alternative hypothesis.

Null hypothesis: μ₁ - μ₂ = 0
Alternative hypothesis: μ₁ - μ₂ ≠ 0

Note that these hypotheses constitute a two-tailed test. The null hypothesis will be rejected if the difference between sample means is too big or if it is too small.

Formulate an analysis plan. For this analysis, the significance level is 0.10. Using sample data, we will conduct a two-sample t-test of the null hypothesis.

Analyze sample data. Using sample data, we compute the standard error (SE), degrees of freedom (DF), and the t statistic test statistic (t).

SE = sqrt[(s₁²/n₁) + (s₂²/n₂)]
SE = 3.109
DF = 10
t = [ (x₁ - x₂) - d ] / SE

t = - 1.61

where s₁ is the standard deviation of sample 1, s₂ is the standard deviation of sample 2, n₁ is the size of sample 1, n₂ is the size of sample 2, x₁ is the mean of sample 1, x₂ is the mean of sample 2, d is the hypothesized difference between the population means, and SE is the standard error.

Since we have a two-tailed test, the P-value is the probability that a t statistic having 10 degrees of freedom is more extreme than -1.61; that is, less than -1.61 or greater than 1.61.

Thus, the P-value = 0.1385

Interpret results. Since the P-value (0.1385) is greater than the significance level (0.10), we cannot accept the null hypothesis.

From the above test we have sufficient evidence in the favor of the claim that there is significance difference on the conversion.