1. Advanced Math Topics 10.2-10.4 Tests Concerning Means for Large Samples

2. Lay’s claims that each bag of potato chips weighs 12 ounces on average, with a standard deviation of 0.8 ounces. A consumer’s group tests this claim by weighing 49 randomly selected bags. The sample mean of the selected bags is 11.8 ounces. Should we reject Lay’s claim? Use a 5% level of significance. Null Hypothesis, H 0 : μ = 12 Alternative Hypothesis, H 1 : μ ≠ 12 μ = 12 Since the alternative hypothesis is that μ ≠ 12, this claim can be shown if the sample mean is much greater OR much less than 12. Thus, it is called a two-tailed test. 0.0250.025 0.4750.475 z = 1.96z = -1.96 If the results of the sample have a z-score greater than 1.96 or less than -1.96, then we can reject Lay’s claim. z = -1.75 Since the z-score is in the acceptance region, our decision is that must accept the claim of Lay’s that their average bag is 12 ounces (H 0 ). The significance level is the probability that the sample mean falls within the rejected region when the null hypothesis is true. Basically, it is the total rejection area for the sample mean. This can be replaced with s, the sample standard deviation. If the sample mean is within our significance level, we will accept the null hypothesis. If the sample mean is outside our significance level, we will reject the null hypothesis and accept the alternate hypothesis.

3. The average score of all 6th graders in a school district on a standardized math test is 75 with a standard deviation of 8.1. A random sample of 100 students at a certain school was taken, and their mean score was 71. Does this indicate that the students of this school are significantly less skilled in math than the average student in the district? Use a 1% level of significance. Null Hypothesis, H 0 : μ = 75 Alternative Hypothesis, H 1 : μ < 75 μ = 75 Since the alternative hypothesis is that μ < 75, this claim can be shown if the sample mean is much less than 75. Thus, it is called a one-tailed test. 0.01 0.49 z = -2.33 If the results of the sample have a z-score less than -2.33, then we can reject the null hypothesis. z = -4.94 Since the z-score is in the rejection region, our decision is that we must reject the null hypothesis. The students at this school are significantly less skilled than the average student in the district. If the sample mean is within our significance level, we will accept the null hypothesis. If the sample mean is outside our significance level, we will reject the null hypothesis and accept the alternate hypothesis.

4. z-scores for Common Significant Levels # of Tails Two-tailed One-tailed 5% Significance Level1% Significance Level z = 1.96z = 2.58 z = 1.645z = 2.33

5. From the HW P. 493 3) The average hourly salary of a statistical typist in a certain city is $14 with a standard deviation of $1.69. The Ajax Typing Service, which employs 55 typist, pays its typists an average hourly rate of $11.88. Can the Ajax Typing Service be accused of paying lower than the average hourly rate? (Use a 5% level of significance.) z = -9.30 Reject the null hypothesis. Ajax pays lower than average hourly rate.

6. HW P. 493 #1-8