various measurements are recorded at different times. bottom temperatures are recorded and a mean temperature of 30.4°C is obtained for the temperatures recorded on 61 different days. σ = 1.7°C, test the claim that the mean population temperature is greater than 30.0°C. Use a 5% level of significance.
i cant find the level of significance chart in my book. i have t and z. am i missing something to find the p-value and test statistic. null and alternate hypothesis.
2006-12-21 04:11:34 · 3 answers · asked by pluca 2 in Science & Mathematics ➔ Mathematics
Depending on how your book sets up the "claim," it will either be the null or alternative hypothesis. Reading the question, it appears to be the alternative hypothesis. Thus, your null hypothesis is that the mean temp is 30 degrees, and the alternative hypothesis is that the mean temp is greater than 30 degrees.
Now you need to find your test statistic. If you are really given sigma (the population standard deviation), you should use a z-test for the mean. x-bar = 30.4, n = 61. Since z = (x-bar - mu)/(sigma/sqrt(n)), z = (30.4 - 30)/(1.7/sqrt(61)) = 1.84.
Since your alternative hypothesis has a greater than sign, your p-value will be the area to the right of your observed test statistic. Thus, p-value = P(Z > 1.84) = 0.0329.
Now you need to compare this p-value to your level of significance, which is kind of like how far away you will tolerate variation according to chance. Since 0.0329 < 0.05, you reject the null hypothesis. You can't stop here, though--make sure to put a conclusion in context of the problem.
2006-12-21 04:25:01 · answer #1 · answered by jbm616 2 · 1⤊ 0⤋
i'm going to ought to come again to you on that. heavily, we studied it this semester. i'm going to ought to check first. I appeared it up. The D-W exams for correlated errors. The attempt statistic is in line with chi-squared and the null hypothesis is uncorrelated errors. As carried out interior the R software, super D-W values help the null meaning no correlation. This propose that the type that yields a DW of .767 is the suitable. because this corresponds to a p-fee of .003, then i might say that it relatively is the suitable type.
2016-10-15 09:20:40 · answer #2 · answered by ? 4 · 0⤊ 0⤋
JBM has it right, just like that.
In another post someone mentioned "If p is low the H0 has got to go" I like that. It means if your area beyond the test statistic (p) is small then you reject H-naught....if p is low the H0 has got to go....hahahaha
2006-12-21 05:40:37 · answer #3 · answered by a_math_guy 5 · 0⤊ 0⤋