Question

9. We want to prove that the average percentage of nitrogen in the air during the Cretaceous era was different than the current 65.1%. The percentage of nitrogen in the air during the Cretaceous era is approximately normally distributed. A random sample of nitrogen percentages from the Cretaceous era is taken (obtained from amber samples) and the data are: 63.4, 65.0, 64.4, 63.3, 54.8, 64.5, 60.8, 49.1, 51.0. Use a = 0.01. a) (7 pts.) Calculate the sample mean and sample standard deviation for the data above. b) (3 pts.) State the appropriate hypotheses. c) (4 pts.) State and calculate the appropriate test statistic. d) (4 pts.) State the rejection region.

Answer 1

Standard Deviation, s: 6.2552866529929 Count, N: 9 Sum, Ex: 536.3 Mean, X: 59.588888888889 Variance, s2: 39.128611111111 Steps N S (x; – T), N 1 i=1 s2 = E(xi - 7) N-1 (63.4 - 59.588888888889)2 + ... +(51 - 59.588888888889) 9-1 313.02888888889 8 = 39.128611111111 S = 139.128611111111 = 6.2552866529929

The provided sample mean is X = 59.589 and the sample standard deviation is 6.255, and the sample size is n = 9. S = (1) Null and Alternative Hypotheses The following null and alternative hypotheses need to be tested: Ho: u = 65.1 Ha: u 7 65.1 This corresponds to a two-tailed test, for which a t-test for one mean, with unknown population standard deviation will be used. (2) Rejection Region Based on the information provided, the significance level is a = 0.01, and the critical value for a two-tailed test is to = 3.355. The rejection region for this two-tailed test is R= {t: 1t| > 3.355} (3) Test Statistics The t-statistic is computed as follows: X — по t= 59.589 – 65.1 6.255/V9 -2.643 s/n (4) Decision about the null hypothesis Since it is observed that |t| = 2.643 <te = 3.355, it is then concluded that the null hypothesis is not rejected. 0.0296 > Using the P-value approach: The p-value is p = 0.0296, and since p= 0.01, it is concluded that the null hypothesis is not rejected. (5) Conclusion It is concluded that the null hypothesis Ho is not rejected. Therefore, there is not enough evidence to claim that the population mean u is different than 65.1, at the 0.01 significance level. Confidence Interval The 99% confidence interval is 52.593 < p < 66.585. Graphically. T-Test Results: t-stats = -2.643, p-value =0.0296 0.40 0.35 0.30 0.25 0.20 0.15 0.10 0.05 0.00 -4.0 -3.5 -3.0 -2.5 -2.0 -1.5 -1.0 -0.5 0.0 0.5 1.0 1.5 2.0 2.5 3.0 3.5 4.0

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