Question

Answer ASAP: Probability and Statistics question

The desired percentage of SiO2 in a certain type of aluminous cement is 5.50 or higher. Suppose that the percentage of SiO2 in a specimen is normally distributed with unknown mean LaTeX: \muμμ but known standard deviation LaTeX: \sigma=0.30σ = 0.30. To test whether the true average percentage is actually less than 5.50 for a particular production facility, we consider

LaTeX: H_0: \mu=5.50 \qquad \textrm{vs} \quad H_1: \mu<5.50H 0 : μ = 5.50 vs H 1 : μ < 5.50

A total of 16 specimens were obtained independently and the corresponding sample mean is LaTeX: \bar{x}=5.25x ¯ = 5.25.

(a) At level 1% can we conclude that the true average percentage is less than 5.50? Make sure to justify your answer.

(b) If the true average percentage is LaTeX: \mu=5.24μ = 5.24 and a level 1% test is used (based on a sample of size 16), what is the power of the test at that LaTeX: \mu μ?

Answer 1

a) As we are testing here whether the mean SiO₂ content is lower than 5.5 on an average, therefore this is a lower tailed test, and the null and the alternative hypothesis here are given as:

The test statistic here is computed as:

As this is a one tailed test, the p-value here is obtained from the standard normal tables as:
p = P(Z < -3.3333) = 0.0004

As the p-value here is 0.0004 < 0.01 which is the level of significance, therefore the test is significant here and we can reject the null hypothesis here. Therefore we have sufficient evidence here that the mean is less than 5.5 here.

b) For 1% level of significance, we have from the standard normal tables,
P( Z < -2.326) = 0.01

The critical mean value here is computed as:

For a true mean value of 5.24, the power of the test is computed as the probability of rejecting the null hypothesis that is computed here as:

Converting it to a standard normal variable, we get here:

Getting it from the standard normal tables, we get here:

therefore 0.873 or 87.3% is the required power of the test here.