3. Aircrew escape systems are powered by a solid propellant. The burning rate of

Question

3. Aircrew escape systems are powered by a solid propellant. The burning rate of this propellant is an important product characteristic. Specifications require that the mean burning rate must be 50 cm/s. The experimenter decides to specify a type I error probability of significance level of = 0.05. He selects a random sample of n = 25 and obtains a sample average burning rate of x_bar = 51.3 cm/s with sample standard deviation of s = 2 cm/s. Measurements indicate that burning rate follows a normal distribution. Test for conformance. What conclusion should he draw?

4. Considering the rocket propellant in problem #3. What is the probability that the test will detect a different true mean (µ’) burning rate if µ’ differs from 50 cm/s by as much as 1 cm/s more than 50 cm/s if the population standard deviation is actually = 1.25 cm/s. (Hint: d = | µ0 – µ’ | / )

Explanation / Answer

Solution:-

3)

State the hypotheses. The first step is to state the null hypothesis and an alternative hypothesis.

Null hypothesis: = 50

Alternative hypothesis: 50

Note that these hypotheses constitute a two-tailed test. The null hypothesis will be rejected if the sample mean is too big or if it is too small.

Formulate an analysis plan. For this analysis, the significance level is 0.05. The test method is a one-sample t-test.

Analyze sample data. Using sample data, we compute the standard error (SE), degrees of freedom (DF), and the t statistic test statistic (t).

SE = s / sqrt(n)

S.E = 0.4

DF = n - 1 = 25 - 1 = 24

t = (x - ) / SE

t = 3.25

where s is the standard deviation of the sample, x is the sample mean, is the hypothesized population mean, and n is the sample size.

Since we have a two-tailed test, the P-value is the probability that the t statistic having 24 degrees of freedom is less than - 3.25 or greater than 3.25

We use the t Distribution Calculator to find P(t < - 3.25) = 0.0006, and P(t > 3.25) = 0.0006

Thus, the P-value = 0.0012

Interpret results. Since the P-value (0.0012) is less than the significance level (0.05), we have to reject the null hypothesis.

From the above test we have sufficient evidence in the favor of a claim that burning rate of this propellant is not equal to 50.

Related Questions

Navigate

• Browse (All)
• Browse 3
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.