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Newly manufactured automobile tires of a certain type are supposed to be filled

ID: 3311214 • Letter: N

Question

Newly manufactured automobile tires of a certain type are supposed to be filled to a pressure of 34 psi. Toavoid over-filling, the manufacturer wants to perform a test on the average pressure filled. To do so, theytake a random sample of 36 tires and average pressure is calculated to be 34.66 psi. Assume that the tirepressures are normally distributed, with standard deviation 2.53 psi.1

. What is the population of interest? What is the parameter of interest for the population? Denote itby.

2. State the null and alternative hypotheses in terms of the parameterabove.

3. Which method would you use?(a) 1-sample T(b) 1-sample Z(c) 2-sample T(d) 2-sample Z

4. Calculate thepvalueof the above testing procedure and make your decision based on= 0.01 in thecontext of the problem.

5. Interpret the abovepvaluein the context.

6. What is the rejection region in terms of the sample average X?

Explanation / Answer

Solution:-

The population of interest is Newly manufactured automobile tires.

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

Null hypothesis: = 34
Alternative hypothesis: 34

Note that these hypotheses constitute a two-tailed test.

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

Analyze sample data. Using sample data, we compute the standard error (SE), z statistic test statistic (z).

SE = s / sqrt(n)

S.E = 0.4217

z = (x - ) / SE

z = 1.57

zcritical = 1.96

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 z statistic having 36 degrees of freedom is less than -1.57 or greater than 1.57.

Thus, the P-value = 0.1164

Interpret results. Since the P-value (0.1164) is greater than the significance level (0.05), we cannot reject the null hypothesis.