Section 10.4: Hypothesis Tests for a Population Mean

Finding P-values for a t test 1.Upper-tailed Test: 1.H a : μ > hypothesized value 2.P-value computed: area in the upper tail 2.Lower-tailed Test: 1.H a : μ < hypothesized value 2.P-value computed: area in the lower tail 3.Two-tailed Test: 1.Ha: μ ≠ hypothesized value 2.P-value computed: sum of area in two tails

One-Sample t Test for a Population Mean Null Hypothesis: H 0 : μ = hypothesized value. Test Statistic:

One-Sample t Test for a Population Mean Alternative Hypothesis: Ha: μ > hypothesized value Ha: μ < hypothesized value Ha: μ ≠ hypothesized value P-value: Area to right of calculated t under t curve with df=n-1 Area to left of calculated t under t curve with df=n-1 (1)2(area to right of t) if t is positive, or (2)2(area to left of t) if t is negative

Assumptions: 1.x bar and s are the sample mean and sample standard deviation, respectively, from a random sample 2.The sample size is large (generally greater than 30) or the population distribution is at least approximately normal.

Example A study conducted by researchers investigated whether time perception, a simple indication of a person’s ability to concentrate, is impaired during nicotine withdrawal. After a 24-hr smoking abstinence, 20 smokers were asked to estimate how much time had passed during a 45-sec period. Suppose the resulting data on perceived elapsed time (in seconds) were as shown (these data are artificial but are consistent with summary quantities given in the paper:

From these data, we obtain n = 20x bar = 59.30s = 9.84

The researchers wanted to determine whether smoking abstinence had a negative impact on time perception, causing elapsed time to be overestimated. We can answer this question by testing H 0 : μ = 45 (no consistent tendency to overestimate the time elapsed) Versus H a : μ > 45 (tendency for elapsed time to be overestimated)

The null hypothesis is rejected only if there is convincing evidence that μ > 45 To answer this question, we carry out a hypothesis test with a significance level of.05 using the step-by-step procedure described in 10.3

1.Population Characteristic of Interest: μ = mean perceived elapsed time for smokers who have abstained from smoking for 24 hrs 2.Null Hypothesis: H 0 : μ = 45 3.Alternative Hypothesis: H a : μ > 45 4.Significance Level: α =.05

6. Assumptions: This test requires a random sample and either a large sample size or a normal population distribution. The authors of the paper believed that it was reasonable to consider this sample as representative of smokers in general, so we regard it as if it were a random sample. Because the sample size is only 20, for the t test to be appropriate, we must be able to assume that the population distribution of perceived elapsed times is at least approximately normal. Is this reasonable? A boxplot shows that there are no outliers so a t test is reasonable.

8. P-value: This is an upper-tailed test (greater than) so the P-value is the area to the right of the computed t value. Because df = 20 – 1 = 19, we can use the df = 19 column of the table to find the p-value. Since it is so high, we find the p-value is ≈ 0.

9. Conclusion: Because p-value ≤ α, we reject H 0 at the.05 level of significance. There is virtually no chance of seeing a sample mean (and hence a t value) this extreme as a result of just chance variation when H 0 is true. There is convincing evidence that the mean perceived time elapsed is greater than the actual time elapsed of 45 sec.