H_0: sigma a = 0.8 alpha = 0.05 n = 16 H_1: sigma 0.8 s = 0.66 4) Determine the
ID: 3149961 • Letter: H
Question
H_0: sigma a = 0.8 alpha = 0.05 n = 16 H_1: sigma 0.8 s = 0.66 4) Determine the number of samples needed in the following situation: a) Detect a shift from n that is plusminus 1 unit (two-sided test) b) sigma is known to be 1.2 units c) The probability of Type 1 error is 0.05 and the probability of Type II error is 0.10 5) You have conducted a study to determine whether your products are meeting the customer requirements. The customer's design states that parts must have a mean length of 220mm. You've collected 8 samples and the mean length of the samples (f) is 218mm. The value of a for part length is 1.1mm.Explanation / Answer
a)
Formulating the null and alternative hypotheses,
Ho: u = 220
Ha: u =/ 220
As we can see, this is a two tailed test.
Thus, getting the critical t,
df = n - 1 = 7
tcrit = +/- 2.364624252
Getting the test statistic, as
X = sample mean = 218
uo = hypothesized mean = 220
n = sample size = 8
s = standard deviation = 1.1
Thus, t = (X - uo) * sqrt(n) / s = -5.142594772
As |t| > 2.635, we REJECT THE NULL HYPOTHESIS.
There is significant evidence at 0.05 level that the true mean length of the parts is not 220 mm. [CONCLUSION]
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b)
Also, the p value is, as this is two tailed,
p = 0.001334898 [ANSWER]
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c)
The probability of making a type I error is alpha, so
P(type I) = alpha = 0.05 [ANSWER]
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d)
A type I error is incorrectly rejecting a true Ho.
Hence, it is incorrectly saying that the true mean length of the parts is not 220 mm, when in fact, it is 220 mm. [ANSWER]
Comparing t and tcrit (or, p and significance level), we REJECT THE NULL HYPOTHESIS.
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