An engineer has designed a valve that will regulate water pressure on an automobile engine. The valve was tested on 200 engines and the mean pressure was 5.4 lbs/square inch. Assume the standard deviation is known to be 0.8. If the valve was designed to produce a mean pressure of 5.5 lbs/square inch, is there sufficient evidence at the 0.05 level that the valve does not perform to the specifications

There is not sufficient evidence at the 0.05 significance level that the valve does not perform to the specifications.

Step-by-step explanation:

Null hypothesis: The valve perform to the specifications.

Alternate hypothesis: The valve does not perform to the specifications.

The population standard deviation is known, hence a z-test is used.

Test statistic (z) = (sample mean - population mean) ÷ (population sd/√n)

sample mean = 5.4

population mean = 5.5

population sd = 0.8

n = 200

z = (5.4 - 5.5) ÷ (0.8/√200) = -0.1 ÷ 0.0566 = -1.77

The test is a two-tailed test. From the standard normal distribution table, the critical values at 0.05 significance level are -1.96 and 1.96.

Decision rules:

Reject the null hypothesis if the test statistic falls outside the region bounded by the critical values.

Fail to reject the null hypothesis if the test statistic falls within the region bounded by the critical values.

Conclusion:

Fail to reject the null hypothesis because the test statistic -1.77 falls within the region bounded by the critical values -1.96 and 1.96.

There is not sufficient evidence at the 0.05 significance level that the valve does not perform to the specifications.

50*10 is greater

step-by-step explanation:

if you add the zero from 10 to 50, it will give you a bigger number.