Please help! I’m lost! I got the x-bar and the standard deviation, now i have to find the t* and p-value?

Use a computer or calculator to complete the calculations and the hypothesis test for this exercise. Delco Products, a division of General Motors, produces commutators designed to be 18.810 mm in overall length. (A commutator is a device used in the electrical system of an automobile.) The following sample of 35 commutators was taken while monitoring the manufacturing process: (The sample does/does not provide sufficient evidence… enter this along with your statement)

18.802 18.810 18.780 18.757 18.824 18.827
18.825 18.809 18.794 18.787 18.884 18.824
18.829 18.817 18.785 18.747 18.802 18.826
18.810 18.802 18.780 18.830 18.874 18.836
18.758 18.813 18.844 18.861 18.824 18.835
18.794 18.853 18.823 18.863 18.808

Is there sufficient evidence to reject the claim that these parts meet the design requirements “mean length is 18.810″ at the = 0.01 level of significance?

Round to 2 places with zero in the units place: t* =

P – value =

Small Sample Hypothesis Test for mean:

In order for this test to be valid the data must come from a normal population. If this is not the case then this test is not valid and other methods, such as a randomization test or permutation test should be used.

Assuming the normality assumption is valid to test the null hypothesis

H0: μ ≤ Δ or
H0: μ ≥ Δ or
H0: μ = Δ
Find the test statistic t = (xbar – Δ ) / (sx / √ (n))

where xbar is the sample average
sx is the sample standard deviation, if you know the population standard deviation, σ , then replace sx with σ in the equation for the test statistic.
n is the sample size

and t follows the Student t distribution with n – 1 degrees of freedom. We use the Student t distribution to account for the uncertainty in the estimate of the variance.
As the degrees of freedom approach infinity the Student t converges in probability to the Standard Normal. In most cases the values of the percentiles of the Student t are close enough to the Standard Normal when the degrees of freedom are greater than 30. This is the source of the empirical rule of thumb that samples of size > 30 have a mean that is normally distributed. Keep that in mind as well, for these hypothesis tests we are assuming the mean is normally distributed. This assumption is easy to verify if the data is normally distributed. The Central Limit Theorem accounts of all other means.

The p-value of the test is the area under the normal curve that is in agreement with the alternate hypothesis.

H1: μ > Δ; p-value is the area to the right of t
H1: μ < Δ; p-value is the area to the left of t
H1: μ ≠ Δ; p-value is the area in the tails greater than |t|

If the p-value is less than or equal to the significance level α, i.e., p-value ≤ α, then we reject the null hypothesis and conclude the alternate hypothesis is true.

If the p-value is greater than the significance level, i.e., p-value > α, then we fail to reject the null hypothesis and conclude that the null is plausible. Note that we can conclude the alternate is true, but we cannot conclude the null is true, only that it is plausible.

The hypothesis test in this question is:

H0: μ = 18.81 vs. H1: μ ≠ 18.81

The test statistic is:
t = ( 18.81534 – 18.81 ) / ( 0.03152025 / √ ( 35 ))
t = 1.002808

The p-value = P( | t_ 34 | > t )
= P( t_ 34 < -1.002808 ) + P( t_ 34 > 1.002808 )
= 2 * P( t_ 34 < -1.002808 )
= 0.3230372

Since the p-value is greater than the significance level of 0.01 we fail to reject the null hypothesis and conclude μ = 18.81 is plausible.

using a "Z" test we have:

Hypothesis Test for mean:

Assuming you have a large enough sample such that the central limit theorem holds, or you have a sample of any size from a normal population with known population standard deviation, then to test the null hypothesis
H0: μ ≤ Δ or
H0: μ ≥ Δ or
H0: μ = Δ
Find the test statistic z = (xbar - Δ ) / (sx / √ (n))

where xbar is the sample average
sx is the sample standard deviation, if you know the population standard deviation, σ , then replace sx with σ in the equation for the test statistic.
n is the sample size

The p-value of the test is the area under the normal curve that is in agreement with the alternate hypothesis.

H1: μ > Δ; p-value is the area to the right of z
H1: μ < Δ; p-value is the area to the left of z
H1: μ ≠ Δ; p-value is the area in the tails greater than |z|

If the p-value is less than or equal to the significance level α, i.e., p-value ≤ α, then we reject the null hypothesis and conclude the alternate hypothesis is true.

If the p-value is greater than the significance level, i.e., p-value > α, then we fail to reject the null hypothesis and conclude that the null is plausible. Note that we can conclude the alternate is true, but we cannot conclude the null is true, only that it is plausible.

The hypothesis test in this question is:

H0: μ = 18.81 vs. H1: μ ≠ 18.81

The test statistic is:
z = ( 18.81534 – 18.81 ) / ( 0.03152025 / √ ( 35 ))
z = 1.002808

The p-value = P( Z > |z| )
= P( Z < -1.002808 ) + P( Z > 1.002808 )
= 2 * P( Z < -1.002808 )
= 0.3159533

Since the p-value is greater than the significance level we fail to reject the null hypothesis and conclude μ = 18.81 is plausible.

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