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8.3 t- test for a mean. T- Test. The t test is a statistical test for the mean of a population and is used when the population is normally or approximately normally distributed and is unknown . The formula for the t test is The degrees of freedom are d.f. = n – 1.

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## 8.3 t- test for a mean

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**T- Test**• The t test is a statistical test for the mean of a population and is used when the population is normally or approximately normally distributedand is unknown. • The formula for the t test is • The degrees of freedom are d.f. = n – 1.**Finding critical values**• Find the critical t value for α = 0.05 with d.f. = 16 for a right-tailed t test.**examples**• 1. Find the critical t value for α = 0.01 with d.f. = 22 for a left-tailed test. • 2. Find the critical value for α = 0.10 with d.f. = 18 for a two-tailed t test.**T-test steps**• Use the same steps as for the z-test. • State null and alternative hypothesis. • Find the critical values from Table F. • Compute the test value. • Reject/Not Reject • Sentence.**Example 1**• A medical investigation claims that the average number of infections per week at a hospital in southwestern Pennsylvania is 16.3. A random sample of 10 weeks had a mean number of 17.7 infections. The sample standard deviation is 1.8. Is there enough evidence to reject the investigator’s claim at α = 0.05? • Step 1: State the hypotheses and identify the claim. H0:μ = 16.3 andH1:μ16.3 • Step 2: Find the critical value. • The critical values are 2.262 and –2.262 for α = 0.05 and d.f. = 9.**Step 3: Find the test value.**• Step 4: Reject/Not • Step 5: Sentence • There is enough evidence to reject the claim that the average number of infections is 16.3.**An educator claims that the average salary of substitute**teachers in school districts in Allegheny County, Pennsylvania, is less than $60 per day. A random sample of eight school districts is selected, and the daily salaries (in dollars) are shown. Is there enough evidence to support the educator’s claim at α = 0.10? 60 56 60 55 70 55 60 55

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