A manufacturing company produces valves in various sizes and shapes. One particular valve plate is supposed to have a tensile strength of 5 pounds per millimeter (lbs/mm). The company tests a random sample of 42 such valve plates and finds a sample mean of 5.0611 lbs/mm and a standard deviation of 0.2803 lbs/mm.a) What is the appropriate null and alternate hypothesis to determine if the average tensile strength of the valve plate is different than 5 lbs/mm?b) Using \alpha= 0.10, what are the critical values for the hypothesis test?c) What is the test statistic for the hypothesis test?d) Draw a conclusion for the hypothesis test. Make sure you state your conclusion in the context of the problem.e) Calculate the p-value for the hypothesis test.

The researcher wants to test if the valve plates manufactured have the expected tensile strength of 5 lbs/mm. So he took a sample of 42 valve plates and measured their tensile strength, obtaining a sample mean of X[bar]= 5.0611 lbs/mm and a sample standard deviation of S=0.2803 lbs/mm.

The study variable is:

X: tensile strength of a valve plate (lbs/mm)

The parameter of interest is the mean tensile strength of the valve plates, μ.

If the claim is that the valve plates of the sample have on average tensile strength of 5 lbs/mm, symbolically: μ = 5

a) The statistic hypotheses are:

H₀: μ = 5

H₁: μ ≠ 5

b) To determine the critical values and rejection region of a hypothesis test you need three to determine three factors of the hypothesis test:

1) The statistical hypothesis.

2) The significance level.

3) The statistic to use for the analysis.

The statistic hypothesis determines the number of critical values and the direction of the rejection region, in this case, the test is two-tailed you will have two critical values and the rejection region will be divided into two.

With the statistic, you will determine the distribution under which you will work and the significance level determines the probability of rejecting the null hypothesis.

To study the population mean you need that the variable of interest has at least a normal distribution, there is no information about the distribution of the study variable but the sample size is large enough n≥30, so you can apply the central limit theorem to approximate the distribution of the sample mean to normal: X[bar]≈N(μ;σ²/n)

Thanks to this approximation it is valid to use an approximation of the standard normal distribution for the test:

[tex]Z= \frac{X[bar]-Mu}{\frac{Sigma}{\sqrt{n} } }[/tex]≈N(0;1)

The critical values are:

[tex]Z_{\alpha /2}= Z_{0.05}= -1.648[/tex]

[tex]Z_{1-\alpha /2}= Z_{0.95}= 1.648[/tex]

You will reject the null hypothesis if [tex]Z_{H_0}[/tex]≤-1.648 or if [tex]Z_{H_0}[/tex]≥1.648

You will not reject the null hypothesis if -1.648<[tex]Z_{H_0}[/tex]<1.648

c)

[tex]Z_{H_0}= \frac{X[bar]-Mu}{\frac{S}{\sqrt{n} } }= \frac{5.0611-5}{\frac{0.2803}{\sqrt{42} } }= 1.41[/tex]

d) The value of the statistic is between the two critical values so the decision is to not reject the null hypothesis. Then using a significance level of 10% there is no significant evidence to reject the null hypothesis so the valve plates have on average tensile strength of 5 lbs/mm.

e) The p-value is defined as the probability corresponding to the calculated statistic if possible under the null hypothesis (i.e. the probability of obtaining a value as extreme as the value of the statistic under the null hypothesis). If the test is two-tailed, so is the p-value, you can calculate it as:

P(Z≤-1.41) + P(Z≥1.41)= P(Z≤-1.41) + (1 - P(Z≤1.41))= 0.079 + ( 1 - 0.921)= 0.158

p-value: 0.158

I hope it helps!