Question

An engineer has designed a valve that will regulate water pressure on an automobile engine. The engineer designed the valve such that it would produce a mean pressure of 7.6 pounds/square inch. The valve was tested on 7 engines and the mean pressure was 7.1 pounds/square inch with a standard deviation of 0.7. Is there evidence at the 0.01 level that the valve does not perform to the specifications? Assume the population distribution is approximately normal.

Step 1 of 5: State the null and alternative hypotheses.

Step 2 of 5:

Find the value of the test statistic. Round your answer to two decimal places.

Step 3 of 5:

Specify if the test is one-tailed or two-tailed.

Step 4 of 5:

Determine the decision rule for rejecting the null hypothesis. Round your answer to three decimal places.

Step 5 of 5:

Make the decision to reject or fail to reject the null hypothesis.

Answer 1

The sample mean is Xˉ=7.1, the sample standard deviation is s=0.7, and the sample size is n=7. (1) Null and Alternative Hypotheses The following null and alternative hypotheses need to be tested: Ho: μ =7.6 Ha: μ ≠7.6 (2)Test Statistics The t-statistic is computed as follows: (3) Two-tailed This corresponds to a Two-tailed test, for which a t-test for one mean, with unknown population standard deviation will be used. (4) Critical Value and Rejection region Based on the information provided, the significance level is α=0.01, and the degree of freedom is n-1=7-1=6. Therefore the critical value for this Two-tailed test is tc​=3.7074. This can be found by either using excel or the t distribution table. Rejection Region The rejection region for this Two-tailed test is therefore: Reject H0 if |t|>3.7074 i.e. t>3.7074 or t<-3.7074 (5) The Decision about the null hypothesis (a) Using traditional method Since it is observed that |t|=1.8898 < tc​=3.7074, it is then concluded that the null hypothesis is Not rejected. (b) Using p-value method The p-value is the probability of obtaining sample results as extreme or more extreme than the sample results obtained, under the assumption that the null hypothesis is true. In this case, the p-value is 0.1077 Using the P-value approach: The p-value is p=0.1077, and since p=0.1077>0.01, it is concluded that the null hypothesis is Not rejected.

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