To test the vendor’s claim regarding the accuracy of the distance measuring instrument, we'll perform a one-sided hypothesis test on the standard deviation, assuming a level of significance ($\alpha$) which is typically set as 0.05 unless specified otherwise. Here's the step-by-step process:

1. Hypotheses Formulation

We aim to test if the instrument’s observed standard deviation significantly exceeds the specified standard deviation (i.e., the vendor's claim).

• Null Hypothesis ($H_0$): The population standard deviation is $\sigma = 0.005 \, \text{m}$.
• Alternative Hypothesis ($H_1$): The population standard deviation is $\sigma > 0.005 \, \text{m}$.

This is a one-sided test since we only want to know if the observed standard deviation is greater than the specified one.

2. Calculation of Sample Standard Deviation

Given the observations: $88.232, \; 88.223, \; 88.222, \; 88.218, \; 88.228, \; 88.215$ we can calculate the sample standard deviation $s$ based on these data points.

3. Test Statistic

To test the standard deviation, we use the chi-square ($\chi^2$) test for variance: $\chi^2 = \frac{(n - 1) \cdot s^2}{\sigma^2}$ where:

• $n = 6$ (sample size),
• $s$ is the sample standard deviation (calculated from the data),
• $\sigma = 0.005 \, \text{m}$ is the population standard deviation under $H_0$.

The degrees of freedom (df) for this test is $n - 1 = 5$.

4. Critical Value and Decision Rule

With a significance level $\alpha = 0.05$, we can find the critical value for $\chi^2$ with 5 degrees of freedom from chi-square distribution tables. We will reject $H_0$ if our calculated $\chi^2$ statistic exceeds this critical value.

5. Calculation and Conclusion

Let's proceed with calculating the sample standard deviation and the test statistic to complete the test.Here are the results of the calculations:

• Sample Standard Deviation ($s$): 0.00626 m
• Chi-square Test Statistic ($\chi^2$): 7.84
• Critical Chi-square Value at $\alpha = 0.05$ and 5 degrees of freedom: 11.07

Conclusion:

Since the calculated test statistic (7.84) is less than the critical chi-square value (11.07), we fail to reject the null hypothesis. This suggests that there is not enough evidence to conclude that the observed standard deviation is greater than the specified value of 0.005 m. Therefore, we approve the vendor's statement regarding the instrument's accuracy at the 5% significance level.

Would you like further details or have questions about this test?

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Here are five questions that extend this information:

1. How would the result change if we had a different level of significance, like 1%?
2. What if the sample size were larger? How would that affect the outcome?
3. How do we interpret the test in the context of the instrument's performance in real-world conditions?
4. Could we use a different test if the sample size were smaller, such as 3 observations?
5. What other factors might we consider in validating the instrument's accuracy?

Tip: In hypothesis testing, the choice of significance level can greatly influence the results; consider the context to set it appropriately.