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7.3 Hypothesis Testing for the Mean (Small Samples)

7.3 Hypothesis Testing for the Mean (Small Samples). Find critical values in a t -distribution Use the t -test to test a mean µ Use technology to find P -values and use them with a t -test to test a mean µ. Find the critical value for a left-tailed test with α = 0.01 and n = 14.

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7.3 Hypothesis Testing for the Mean (Small Samples)

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  1. 7.3 Hypothesis Testing for the Mean (Small Samples) Find critical values in a t-distribution Use the t-test to test a mean µ Use technology to find P-values and use them with a t-test to test a mean µ

  2. Find the critical value for a left-tailed test with α = 0.01 and n = 14. Critical value: -2.650 Try it yourself 1 Finding Critical Values for t

  3. Find the critical value for a right-tailed test with α = 0.10 and n = 9. Critical value: 1.397 Try it yourself 2 Finding Critical Values for t

  4. Find the critical values –t and t for a two-tailed test with α = 0.05 and n = 16. Critical values: -2.131 and 2.131 Try it yourself 3 Finding Critical Values for t

  5. The t-test for a mean is a statistical test for a population mean. The t-test can be used when the population is normal or nearly normal, σ is unknown, and n < 30. The test statistic is the sample mean and the standardized test statistic is t-Test for a mean µ

  6. An insurance agent says that the mean cost of insuring a 2008 Honda CR-V is less than $1200. A random sample of 7 similar insurance quotes has a mean cost of $1125 and a standard deviation of $55. Is there enough evidence to support the agent’s claim at α = 0.10? Assume the population is normally distributed. Critical value: -1.440; Rejection region: t < -1.440 t = -3.61 Reject the null hypothesis There is enough evidence at the 10% level of significance to support the claim that the mean cost of insuring a 2008 Honda CR-V is less than $1200. Try it yourself 4 Testing µ with a Small Sample

  7. An industrial company claims that the mean conductivity of the water in a nearby river is 1890 milligrams per liter. The conductivity of a water sample is a measure of the total dissolved solids in the sample. You randomly select 19 water samples and measure the conductivity of each. The sample mean and standard deviation are 2500 milligrams per liter and 700 milligrams per liter, respectively. Is there enough evidence to reject the company’s claim at α = 0.01? Assume the population is normally distributed. Critical values: -2.878 and 2.878 Rejection regions: t < -2.878, t > 2.878 t = 3.798 Reject the null hypothesis There is enough evidence at the 1% level of significance to reject the claim that the mean conductivity of the river is 1890 milligrams per liter. Try it yourself 5 Testing µ with a Small Sample

  8. A Department of Motor Vehicles office claims that the mean wait time is at most 18 minutes. A random sample of 12 people has a mean wait time of 15 minutes with a standard deviation of 2.2 minutes. At α = 0.05, test the office’s claim. Assume the population is normally distributed. P = 0.9997 Fail to reject the null hypothesis. There is not enough evidence at the 5% level of significance to reject the claim that the mean wait time is at most 18 minutes. Try it yourself 6 Using P-values with a t-test

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