We appreciate your visit to An engineer has designed a valve that will regulate water pressure on an automobile engine The valve was tested on 160 engines and the mean. This page offers clear insights and highlights the essential aspects of the topic. Our goal is to provide a helpful and engaging learning experience. Explore the content and find the answers you need!
Answer :
t = (sample mean - hypothesized mean) / (standard deviation / square root of sample size)
Substituting the values given in the question, we get:
t = (6.2 - 6) / (0.9 / sqrt(160))
t = 5.46
Since the p-value is less than the significance level of 0.05, we can reject the null hypothesis and conclude that there is sufficient evidence at the 0.05 level that the valve performs above the specifications.
To determine if there is sufficient evidence at the 0.05 level that the valve performs above the specifications, we will perform a hypothesis test using the given information.
Step 1: State the null and alternative hypotheses.
Null hypothesis (H0): The mean water pressure is equal to 6 lbs/square inch (µ = 6).
Alternative hypothesis (H1): The mean water pressure is greater than 6 lbs/square inch (µ > 6).
The null hypothesis (H0) is that the mean pressure produced by the valve is 6 lbs/square inch, and the alternative hypothesis (Ha) is that the mean pressure produced by the valve is greater than 6 lbs/square inch.
We can use a one-sample t-test to test this hypothesis.
The test statistic is calculated as:
t = (sample mean - hypothesized mean) / (standard deviation / square root of sample size)
Step 2: Calculate the test statistic.
The test statistic (z) = (sample mean - population mean) / (standard deviation/sqrt (sample size))
z = (6.2 - 6) / (0.9 / sqrt(160))
z ≈ 3.11
Step 3: Determine the critical value and make a decision.
Using a t-distribution table with 159 degrees of freedom (160-1), we find that the probability of getting a t-value of 5.46 or greater is very small, less than 0.0001.
Since this is a one-tailed test at the 0.05 significance level, we will compare the test statistic (z) to the critical value from the z-table. The critical value for a one-tailed test at a 0.05 significance level is 1.645.
Since the test statistic (3.11) is greater than the critical value (1.645), we reject the null hypothesis in favor of the alternative hypothesis.
Conclusion: There is sufficient evidence at the 0.05 level that the valve performs above the specifications, as the mean water pressure is greater than 6 lbs/square inch.
Learn more about the null hypothesis:
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