Two Population Means with Known Standard Deviations Quiz

PDF page 440: Even though this situation is not likely (knowing the population standard deviations is very unlikely), the following example illustrates hypothesis testing for independent means with known population standard deviations. The sampling distribution for the difference between the means is normal in accordance with the central limit theorem. The random variable is – – X – X . The normal distribution has the following format: 1 2 The standard deviation is: 2 2

PDF page 440: Even though this situation is not likely (knowing the population standard deviations is very unlikely), the following example illustrates hypothesis testing for independent means with known population standard deviations. The sampling distribution for the difference between the means is normal in accordance with the central limit theorem. The random variable is – – X – X . The normal distribution has the following format: 1 2 The standard deviation is: 2 2

PDF page 440: Even though this situation is not likely (knowing the population standard deviations is very unlikely), the following example illustrates hypothesis testing for independent means with known population standard deviations. The sampling distribution for the difference between the means is normal in accordance with the central limit theorem. The random variable is – – X – X . The normal distribution has the following format: 1 2 The standard deviation is: 2 2

PDF page 441: The test statistic (z-score) is: – – ( x – x ) – δ

PDF page 441: 1 2 0 Zc = 2 2 (σ ) (σ ) 2 1

PDF page 441: Independent groups, population standard deviations known: The mean lasting time of two competing floor waxes is to be compared. Twenty floors are randomly assigned to test each wax. Both populations have a normal distributions. The data are recorded in Table 10.3.

PDF page 441: Independent groups, population standard deviations known: The mean lasting time of two competing floor waxes is to be compared. Twenty floors are randomly assigned to test each wax. Both populations have a normal distributions. The data are recorded in Table 10.3.

PDF page 441: Does the data indicate that wax 1 is more effective than wax 2? Test at a 5% level of significance.

PDF page 441: Solution 10.7 This is a test of two independent groups, two population means, population standard deviations known. – – Random Variable: X – X = difference in the mean number of months the competing floor waxes last. 1 2

PDF page 441: 1 2 The words "is more effective" says that wax 1 lasts longer than wax 2, on average. "Longer" is a “>” symbol and goes into Ha. Therefore, this is a right-tailed test. Distribution for the test: The population standard deviations are known so the distribution is normal. Using the formula for the test statistic we find the calculated value for the problem.

PDF page 442: The estimated difference between he two means is : X – X = 3 – 2.9 = 0.1

PDF page 442: 1 2 Compare calculated value and critical value and Zα: We mark the calculated value on the graph and find the the calculate value is not in the tail therefore we cannot reject the null hypothesis. Make a decision: the calculated value of the test statistic is not in the tail, therefore you cannot reject H0. Conclusion: At the 5% level of significance, from the sample data, there is not sufficient evidence to conclude that the mean time wax 1 lasts is longer (wax 1 is more effective) than the mean time wax 2 lasts.

PDF page 442: are randomly assigned to be tested. Both populations have normal distributions. Table 10.4 shows the result. Do the data indicate that Engine 2 has higher RPM than Engine 1? Test at a 5% level of significance.