Question

The desired percentage of SiO₂ in a certain type of aluminous cement is 5.5. To test whether the true average percentage is 5.5 for a particular production facility, 16 independently obtained samples are analyzed. Suppose that the percentage of SiO₂ in a sample is normally distributed with ? = 0.32 and that x = 5.21. (Use ? = 0.05.)

(a) Does this indicate conclusively that the true average percentage differs from 5.5?
State the appropriate null and alternative hypotheses.

H₀: ? = 5.5
H_a: ? > 5.5H₀: ? = 5.5
H_a: ? < 5.5     H₀: ? = 5.5
H_a: ? ? 5.5H₀: ? = 5.5
H_a: ? ? 5.5

Calculate the test statistic and determine the P-value. (Round your test statistic to two decimal places and your P-value to four decimal places.)

z	=
P-value	=

State the conclusion in the problem context.

Do not reject the null hypothesis. There is not sufficient evidence to conclude that the true average percentage differs from the desired percentage.Reject the null hypothesis. There is sufficient evidence to conclude that the true average percentage differs from the desired percentage.     Do not reject the null hypothesis. There is sufficient evidence to conclude that the true average percentage differs from the desired percentage.Reject the null hypothesis. There is not sufficient evidence to conclude that the true average percentage differs from the desired percentage.

(b) If the true average percentage is

? = 5.6

and a level ? = 0.01 test based on n = 16 is used, what is the probability of detecting this departure from H₀? (Round your answer to four decimal places.)


(c) What value of n is required to satisfy ? = 0.01 and ?(5.6) = 0.01? (Round your answer up to the next whole number.)
n =  samples

Answer 1

Solution:

(a) The null and alternative hypotheses are, HO: 5.5 Here, μ-μ0 is the hypothesized mean. Let the sample mean be x- 5.21 Let the population standard deviation be ơ-0.32. Let the number of independent samples be n-16 The test statistic value is, z0 (5.21-5.5) 0.32/16) -3.625 -3.63 The two tailed p-value is, p-2P(z < -3.63) 2(0.0001)- 0.0002 Let the level of significance be α 0.05 The p-value is less than the significance level of 0.05. Reject the nul hypothesis. There is sufficient evidence to conclude that the true average percentage differs from the desired percentage (b) Let α-001 and β-0.01 Using the standard normal distribution area tables, the critical value at 0.01 level is 2.576 Let the tre population mean be μ,-5.6 The type II error for a two tailed hypothesis Ha: μ μ0 is. Φ(2.576 + 55-56/32N16)-$(-2.576 + 55-5632N16) 0.9076- 0.0001 0.9075 (c) The sample size n for which a level a o0.01 test also has B(5.6)0.01 at the alternative value -5.6 is, 0.32(2.576+ 2.576)/5.5-5.6]*2 - 271.8014 272 samples