B1 hypothesis test calculator how to#
There is insufficient evidence, at the \(\alpha\) = 0.05 level, to conclude that the mean thickness of all pieces of spearmint gum differs from 7.5 one-hundredths of an inch.We now show how to test the value of the slope of the regression line. Since the P-value, 0.158, is greater than \(\alpha\) = 0.05, the quality control specialist fails to reject the null hypothesis. In the output above, Minitab reports that the P-value is 0.158. If the quality control specialist used the P-value approach to conduct his hypothesis test, he would determine the area under a t n - 1 = t 9 curve, to the right of 1.54 and to the left of -1.54: That is, the test statistic does not fall in the "critical region." There is insufficient evidence, at the \(\alpha\) = 0.05 level, to conclude that the mean thickness of all of the manufacturer's spearmint gum differs from 7.5 one-hundredths of an inch.
Since the quality control specialist's test statistic, t* = 1.54, is not less than -2.2616 nor greater than 2.2616, the quality control specialist fails to reject the null hypothesis.
B1 hypothesis test calculator software#
If the quality control specialist sets his significance level \(\alpha\) at 0.05 and used the critical value approach to conduct his hypothesis test, he would reject the null hypothesis if his test statistic t* were less than -2.2616 or greater than 2.2616 (determined using statistical software or a t-table): The test statistic t* is 1.54, and the P-value is 0.158. (The standard error of the mean "SE Mean", calculated by dividing the standard deviation 0.1027 by the square root of n = 10, is 0.0325). The output tells us that the average thickness of the n = 10 pieces of gums was 7.55 one-hundredths of an inch with a standard deviation of 0.1027.
There is sufficient evidence, at the \(\alpha\) = 0.05 level, to conclude that the mean height of all such sunflower seedlings is less than 15.7 cm.Īlternative hypothesis H₁: $\mu \ne$ 7.5 T-Value Since the P-value is less than 0.001, it is clearly less than \(\alpha\) = 0.05, and the biologist rejects the null hypothesis. In the output above, Minitab reports that the P-value is 0.000, which we take to mean < 0.001. If the biologist used the P-value approach to conduct her hypothesis test, she would determine the area under a t n - 1 = t 32 curve and to the left of the test statistic t* = -4.60: That is, the test statistic falls in the "critical region." There is sufficient evidence, at the α = 0.05 level, to conclude that the mean height of all such sunflower seedlings is less than 15.7 cm. Since the biologist's test statistic, t* = -4.60, is less than -1.6939, the biologist rejects the null hypothesis. If the biologist set her significance level \(\alpha\) at 0.05 and used the critical value approach to conduct her hypothesis test, she would reject the null hypothesis if her test statistic t* were less than -1.6939 (determined using statistical software or a t-table):s-3-3 Throughout this course (and your future research!), when you see that Minitab reports the P-value as 0.000, you should report the P-value as being "< 0.001." If Minitab reports the P-value as 0.000, it really means that the P-value is 0.000.something. Minitab will always report P-values to only 3 decimal places.
There is insufficient evidence, at the \(\alpha\) = 0.05 level, to conclude that the mean Brinell hardness of all such ductile iron pieces is greater than 170. Since the P-value, 0.117, is greater than \(\alpha\) = 0.05, the engineer fails to reject the null hypothesis. In the output above, Minitab reports that the P-value is 0.117. If the engineer used the P-value approach to conduct his hypothesis test, he would determine the area under a t n - 1 = t 24 curve and to the right of the test statistic t* = 1.22: That is, the test statistic does not fall in the "critical region." There is insufficient evidence, at the \(\alpha\) = 0.05 level, to conclude that the mean Brinell hardness of all such ductile iron pieces is greater than 170. Since the engineer's test statistic, t* = 1.22, is not greater than 1.7109, the engineer fails to reject the null hypothesis. If the engineer set his significance level α at 0.05 and used the critical value approach to conduct his hypothesis test, he would reject the null hypothesis if his test statistic t* were greater than 1.7109 (determined using statistical software or a t-table): The test statistic t* is 1.22, and the P-value is 0.117.
(The standard error of the mean "SE Mean", calculated by dividing the standard deviation 10.31 by the square root of n = 25, is 2.06). The output tells us that the average Brinell hardness of the n = 25 pieces of ductile iron was 172.52 with a standard deviation of 10.31. Alternative hypothesis H₁: $\mu$ > 170 T-Value
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