The diameter of a shaft assembled on a specific component of a space shuttle is made of titanium and must have a tensile strength of at least 97.5 psi. From previous quality inspections, it is known that the standard deviation of tensile strength is 10.4 psi. Quality engineers took a sample of 25 shafts and measured the tensile strength, finding a sample average of [tex]x = 103 \, \text{psi}[/tex].

Quality engineers want to be certain that the shafts will have at least 97.5 psi tensile strength. Prepare a hypothesis test, including the test statistic and P-value.

1. State the Hypotheses:
- Null Hypothesis ([tex]H_0[/tex]): The mean tensile strength is at least 97.5 psi ([tex]\mu \geq 97.5[/tex]).
- Alternative Hypothesis ([tex]H_a[/tex]): The mean tensile strength is less than 97.5 psi ([tex]\mu < 97.5[/tex]).

2. Test Statistic:
- Use the formula for the test statistic:
\[
z = \frac{\bar{x} - \mu_0}{\sigma/\sqrt{n}}
\]
where:
- [tex]\bar{x} = 103[/tex] psi (sample mean)
- [tex]\mu_0 = 97.5[/tex] psi (hypothesized population mean)
- [tex]\sigma = 10.4[/tex] psi (population standard deviation)
- [tex]n = 25[/tex] (sample size)

3. P-Value:
- Calculate the P-value using the test statistic.

Note: The completion of the calculation and interpretation of the P-value will depend on statistical software or Z-table references.

The hypothesis test conducted suggests that the population mean tensile strength of the titanium shafts is greater than 97.5 psi, based on the sample data. With a test statistic of 2.644 and a sufficiently low p-value, the null hypothesis can be rejected, indicating the shafts meet the strength requirement.

Explanation:

To test the hypothesis that the shafts have at least 97.5 psi tensile strength, we set up the following null (H0) and alternative (Ha) hypotheses:

H0: μ = 97.5 psi (The population mean tensile strength is equal to 97.5 psi)
Ha: μ > 97.5 psi (The population mean tensile strength is greater than 97.5 psi)

We will use a one-sample t-test since the population standard deviation is known and the sample size is small (n=25). The test statistic is calculated as follows:

t = (x - μ) / (s/√n)

Plugging in the values, we get:

t = (103 - 97.5) / (10.4/√25) = 5.5 / (10.4/5) = 2.644

Next, we calculate the p-value using a t-distribution table or software for a one-tailed test. Given the high value of t and a degree of freedom (df) of 24, the p-value will be very low, much below the alpha level of 0.05, indicating strong evidence against the null hypothesis.