Hypothesis Testing - sigma known. Use Z distribution. The manager of aircraft ma
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Question
Hypothesis Testing - sigma known. Use Z distribution. The manager of aircraft maintenance for an airline wants to test whether the monthly maintenance cost of aircraft equals the projected amount of $600 per plane. The population standard deviation sigma is known to be $100. A sample of 36 planes is taken, which shows a mean cost of $680. Test whether the cost is different from the projected amount. That is: Test: H_0: mu = $600 versus H_A: mu notequalto $600. Use alpha = 05. Do both of these: Compute the test statistic Z = X bar - mu_0/sigma/squareroot n & compare it to the critical value from the Z-table. Compute the upper and lower values of X bar: (X bar*_1, X bar*_2), which if X bar is outside that range would cause you to reject H_0. Hypothesis Testing - sigma known. Use Z distribution. A company that produces snack foods uses a machine that packages 454 gram bags of pretzels. The weights are normally distributed and the population standard deviation is known to be 7.8 grams. A sample of 25 bags of pretzels is taken, and the sample has a mean weight of 450 grams. Test whether the weights are different from the promised amount, or in other words whether the machine is working properly. That is: Test: H_0: mu = 454 versus H_A: mu notequalto 454. Use alpha = 05. Do both of these: Compute the test statistic t = X bar - mu_0/S/squareroot n & compare it to the critical value from the Z-table. Compute the upper and lower values of X bar: (X bar*_1, X bar*_2), which if X bar is outside that range would cause you to reject H_0. P - values An aircraft manufacturer needs specialty steel rods with a tensile strength of at least 5,000 pounds. The firm's quality inspector tests a random sample of incoming steel rods. 64 rods are tested (n = 64). The standard deviation is 800 pounds. (Assume that this represents the known population standard deviation sigma). The inspector finds a mean strength of X bar = 4800 pounds. Test H_0: mu greaterthanorequalto 5000 pounds versus H_A: muExplanation / Answer
Solution:-
State the hypotheses. The first step is to state the null hypothesis and an alternative hypothesis.
Null hypothesis: = 600
Alternative hypothesis: 600
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 z-test.
Analyze sample data. Using sample data, we compute the standard error (SE), and the z statistic test statistic (z).
SE = s / sqrt(n)
S.E = 16.667
z = (x - ) / SE
z = 4.8
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 statistics less than - 4.8 or greater than 4.8.
Thus, the P-value = 0.000029
Interpret results. Since the P-value (0.000029) is less than the significance level (0.05), we have to reject the null hypothesis.
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