1.
t0.01 = 2.602
-t0.05 = -1.729
plus or minus t0.025 = plus or minus 2.179
2.
1.321
-2.426
plus or minus 2.738
3.
t0 = -1.379
-t0.05 = -1.714
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There isn’t enough evidence for the researcher to reject the null hypothesis because it is a left-tailed test and the test statistic is greater than the critical value. (-1.370 > -1.714)
4.
t0 = 2.674
-t0.1 = -1.318, t0.1 = 1.318
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No, because there is sufficient amount of evidence.
5.
t0 = 2.502
-t0.005 = -2.819; t0.005 = 2.819
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There is not sufficient evidence for the researcher to reject the null hypothesis, since the test statistic is between the critical values. (-2.819 < 2.502 < 2.819)
Lower bound: 99.39, Upper bound: 110.21. Because the 99% confidence interval includes 100, we do not reject the statement in the null hypothesis.
11.
Ho: mew = $67
H1: mew > $67
There is a 0.02 probability of obtaining a sample mean of $73 or higher from a population whose mean is $67. So, if we obtained 100 simple random sampldes of size n = 40 from a populationwhose mean is $67, we sould expect about twoof these samples to result in sample means of $73 or higher.
Vecause the P-value is low (p-value = 0.02 < beta = 0.05), we reject the statement in the null hypothesis. There’s sufficient evidenceto conclude that the mean dollar amount withdrawn from a PayEaste ATM is more than the mean amount from a standard ATM.
13.
Ho: mew = 22
H1: mew > 22
The sample is random. The sample size is large, n = 200 greater than or equal to 30. We can reasonably assume that the sample is small relative to the population, so the scores are independent.
Classical approah: t0 = 2.176 > t0.05 = 1.660; reject the null hypothesis. P-value approach: 0.01 < p-value < 0.02; reject the null hypothesis.
There is sufficient evidence to conclude that students who complete the core curriculum are scoring above 22 on the math portion of the ACT.
15.
P-value < 0.0005 < beta = 0.01; rejects the null hypothesis. There is sufficient evidence to conclude that the mean hippocampal volume in alchoholic adolescents is less than thenormal mean volume of 0.02 cm^3.
17.
0.25 > P-value > 0.20 > beta = 0.05; do not reject the null hypothesis. There is not sufficient evidence; it is not unlikely to obtain a mean credit score of 714.2 or higher even though the true population mean credit score is 703.5.
19.
Lower bound: 35.44 years, Upper bound: 42.36 years. Because the interal includes 40.7 years, there is not sufficient evidence to conclude that the mean age of a death-row inmate has changed since 2002.
1.
chi^2 0.05 = 28.869
chi^2 0.9 = 14.041
chi^2 0.975 = 16.047; chi^2 0.025 = 45.722
3.
chi^2 0 = 20.496
chi^2 0.95 = 13.091
Skip but feel free to sketch on a piece of paper for your benefit.
Do not reject H0, because chi^2 0 > chi^2 0.95.
13. Feel free to use any method you want. If you use the classical method please report the value of your test statistic and your decision. Likewise if you use the pvalue method please state your pvalue and your decision. Hint: The sample standard deviation (s) is 15.20.
0.10 < p-value < 0.90; do not reject the null hypothesis at the beta = 0.05 level of significance. There is not sufficient evidence at the beta = 0.05 level of significance. There is not sufficient evidence at the beta = 0.05 level of significance to conclude that the standard deviation wait-time is less than 18 seconds.
15. Feel free to use any method you want. If you use the classical method please report the value of the test statistic and your decision. Likewise if you use the pvalue method please state your pvalue and your decision.
0.05 < p-value < 0.10 < beta = 0.10; reject the null hypothesis at the beta = 0.10 level of significance. There is sufficient evidence at the beta = 0.10 level of significance to conclude that Rose is a more consistent player than other shooting guards in the NBA.