In: Statistics and Probability
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 σ = 0.32 and that x = 5.22. (Use α = 0.05.)
(a) Does this indicate conclusively that the true average
percentage differs from 5.5?
State the appropriate null and alternative hypotheses.
1)H0: μ = 5.5
Ha: μ < 5.5
2)H0: μ = 5.5
Ha: μ ≥
5.5
3)H0: μ = 5.5
Ha: μ ≠ 5.5
4)H0: μ = 5.5
Ha: μ > 5.5
b)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 | = |
c)State the conclusion in the problem context.
1)Reject the null hypothesis. There is sufficient evidence to conclude that the true average percentage differs from the desired percentage.
2)Do not reject the null hypothesis. There is not sufficient evidence to conclude that the true average percentage differs from the desired percentage.
3)Do not reject the null hypothesis. There is sufficient evidence to conclude that the true average percentage differs from the desired percentage.
4)Reject the null hypothesis. There is not sufficient evidence to conclude that the true average percentage differs from the desired percentage.
(d) If the true average percentage is μ = 5.6 and a level
α = 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.)
(e) What value of n is required to satisfy α =
0.01 and β(5.6) = 0.01? (Round your answer up to the next
whole number.)
n = ______ samples
(a)
The appropriate hypothesis are,
3)H0: μ = 5.5
Ha: μ ≠ 5.5
(b)
Standard error of mean = = 0.32 / = 0.08
Test statistic, z = (5.22 - 5.5) / 0.08 = -3.5
For two tail test, p-value = 2 * P(z < -3.5) = 0.0005
(c)
Since p-value is less than the significance level of 0.05,
1)Reject the null hypothesis. There is sufficient evidence to conclude that the true average percentage differs from the desired percentage.
(d)
For α = 0.01 , critical value of z is -2.576 and 2.576
Critical values of x to reject H0 = 5.5 - 0.08 * 2.576 = 5.294 and 5.5 + 0.08 * 2.576 = 5.706
That is , we reject H0 if x < 5.294 and x > 5.706
Probability of detecting this departure from H0 = P(x < 5.294) + P(x > 5.706)
= P[z < (5.294 - 5.6) / 0.08] + P[z > (5.706 - 5.6) / 0.08]
= P[z < -3.825] + P[z > 1.325]
= 0.0927
(e)
For α = 0.01 , critical value of z is -2.576 and 2.576
Critical values of x to reject H0 = 5.5 - 0.32/ * 2.576 and 5.5 + 0.32/ * 2.576
= 5.5 - 0.82432 / , 5.5 + 0.82432 /
β(5.6) = 0.01
=> P(5.5 - 0.82432 / < x < 5.5 + 0.82432 / ) = 0.01
Because of symmetry of normal distribution,
P(x < 5.5 + 0.82432 / ) = 0.5 + 0.01/2 = 0.505
P[z < (5.5 + 0.82432 / - 5.6) / (0.32 / )] = 0.505
By Z distribution table,
(-0.1 + 0.82432 / ) / (0.32 / ) = 0.0125
-0.1 + 0.82432 / = 0.004 /
(0.82432 - 0.004) / = 0.1
n = (0.82032 / 0.1)2 = 68 (Rounding to the next whole number.)