Making Decisions about a Population Mean with Confidence Lecture 33 Sections 10 1 10 2 Tue Oct 30 2007 Introduction In Chapter 10 we will ask the same basic questions as in Chapter 9 except they will concern the mean Find an estimate for the mean Test a hypothesis about the mean The Steps of Testing a Hypothesis p Value Approach 1 State the null and alternative hypotheses 2 State the significance level 3 Give the test statistic including the formula 4 Compute the value of the test statistic 5 Compute the p value 6 State the decision 7 State the conclusion The Hypotheses The null and alternative hypotheses will be statements concerning Null hypothesis H0 0 Alternative hypothesis choose one 0 H1 0 H1 0 H1 The Hypotheses See Example 10 1 p 616 The hypotheses are H0 15 mg H1 15 mg Level of Significance The level of significance is the same as before If the value is not given assume that 0 05 The Test Statistic The choice of test statistic will depend on the sample size and what is known about the population Details to follow If we assume that is known and that either The sample size n is at least 30 or The population is normal Then the Central Limit Theorem for Means will apply See p 615 The Sampling Distribution of x If the population is normal then the distribution of x is also normal with mean 0 and standard deviation n x is exactly N 0 n for any sample size no matter how small This assumes that is known The Sampling Distribution of x Therefore the test statistic is x 0 Z n It is exactly standard normal The Sampling Distribution of x On the other hand if The population is not normal But the sample size is at least 30 then the distribution of x is approximately normal with mean 0 and standard deviation n x is approximately N 0 n We are still assuming that is known The Sampling Distribution of x Therefore the test statistic is x 0 Z n It is approximately standard normal The approximation is good enough that we can use normalcdf The Decision Tree yes Is known no The Decision Tree yes Is the population normal yes no Is known no The Decision Tree yes Is the population normal yes no X Z n Is known no The Decision Tree Is known yes Is the population normal yes no X Z n Is n 30 yes no no The Decision Tree Is known yes Is the population normal yes no X Z n Is n 30 yes X Z n no no The Decision Tree Is known yes Is the population normal yes no X Z n Is n 30 yes X Z n no Give up no The Decision Tree Is known yes Is the population normal yes no X Z n Come back tomorrow Is n 30 yes X Z n no no Give up Calculate the Value of the Test Statistic In our sample we find that x 12 528 We are assuming that 4 8 Therefore 12 528 15 2 472 Z 2 575 0 96 4 8 25 Compute the p Value The p value is P x 2 575 Use normalcdf E99 2 575 0 005012 Therefore p value 0 005012 Decision Because the p value is less than we will reject the null hypothesis Conclusion We conclude that the carbon monoxide content of cigarettes is lower today than it was in the past Hypothesis Testing on the TI 83 Press STAT Select TESTS Select Z Test Press ENTER A window appears requesting information Select Data if you have the sample data entered into a list Otherwise select Stats Hypothesis Testing on the TI 83 Assuming you selected Stats Enter 0 the hypothetical mean Enter Remember is known Enter x Enter n the sample size Select the type of alternative hypothesis Select Calculate and press ENTER Hypothesis Testing on the TI 83 A window appears with the following information The title Z Test The alternative hypothesis The value of the test statistic Z The p value of the test The sample mean The sample size Example Re do Example 10 1 on the TI 83 using Stats The TI 83 reports that z 2 575 p value 0 005012 Hypothesis Testing on the TI 83 Suppose we had selected Data instead of Stats Then somewhat different information is requested Enter the hypothetical mean Enter Why Identify the list that contains the data Skip Freq it should be 1 Select the alternative hypothesis Select Calculate and press ENTER Example Re do Example 10 1 on the TI 83 using Data Enter the data in the chart on page 616 into list L1 The TI 83 reports that z 2 575 p value 0 005012 x 12 528 s 4 740 4 8
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