Answer:
a. You cannot conclude that the true mean of the inside diameter is different from 1.5 inches.
b. p-value: 0.992
Step-by-step explanation:
Hello!
To test the hypothesis that "The true mean of inside diameter is different from 1.5 inches."
Let's define the study variable:
X: "inside diameter of a bearing"
X~N(μ;[tex]σ^{2}[/tex])
With known standart derivation σ= 1 inch
Sample taken:
n= 25 bearings
X(bar)= 1.4975 inches
>Statistical hyphotesis for this problem
H₀: μ = 1.5
H₁: μ ≠ 1.5
>Level of significance
α: 0,01
>Since the variable has a normal distribution and the population standard desviation is known the mos accurate statistical to use is:
[tex]Z\frac{x(bar)-μ}{\frac{σ}{\sqrt[]{n} } }[/tex]
>The critical region for this test is two-tailed, that means you'll have two critical values.
[tex]Z_{0,005} = -2,58[/tex]
[tex]Z_{0,995} = 2,58[/tex]
This means, you'll reject the null hyphotesis if your [tex]Z_{obs}[/tex] is ≤ -2,58 or if it is ≥ 2,58
Now calculate the [tex]Z_{obs}[/tex] value by replacing the formula:
[tex]Z\frac{x(bar)-μ}{\frac{σ}{\sqrt[]{n} } } = \frac{1.4975-1.5}{\frac{1}{\sqrt[]{25} } } = \frac{-0.0025}{0.2} = -0.0125[/tex]
Since the calculated Z value is between the non rejection area, the test is not statistically significant. In other words, you cannot conclude that the true mean of the inside diameter is diferent from 1.5 inches.
b. To calculate the p-value, you have to look for the probability of the calculated Z-value
P(Z≤-0.0125) ≅ 0,496
The p-value above is one tailed, to have the p-value for this thest you need to simply multiply it by 2, so the p-value for the test is 0.992 (aprox)
