Learn Significance Test with Examples & Questions

Recent questions in Significance tests
College StatisticsAnswered question
vrotterigzl vrotterigzl 2022-06-26

I am doing a two sample hypothesis problem that goes like this:
Research has shown that good hip range of motion and strength in throwing athletes results in improved performance and decreased body stress. The article “Functional Hip Characteristics of Baseball Pitchers and Position Players” (Am. J. Sport. Med., 2010: 383–388) reported on a study involving samples of 40 professional pitchers and 40 professional position players. For the pitchers, the sample mean trail leg total arc of motion (degrees) was 75.6 with a sample standard deviation of 5.9, whereas the sample mean and sample standard deviation for position players were 79.6 and 7.6, respectively. Assuming normality, test appropriate hypotheses to decide whether true average range of motion for the pitchers is less than that for the position players (as hypothesized by the investigators).
Using the information from the question I was able to get a t value of −2.63, from what I understand from here I'm supposed to compare this -2.63 to a critical value I get from the t table, or I can get find the p-value (I got 0.0043 from the normal table) and compare it to alpha which is the significance level. For both of these I don't really understand on which situations I'm supposed to be using for which method, It would be great if someone could explain this to me too. But regardless the main issue I have is that both of these requires a significance level to figure out, which wasn't given to me. Since I didn't know how to proceed I went to look at the answer to this question, which said this:
Because the one-tailed P-value is .005<.01, conclude at the .01 level that the difference is as stated.
The alternative hypothesis was chosen because the P value was lower than the significance level of .01 but if they had just chosen a smaller significance level(e.g. 0.001), then the alternative hypothesis would have been rejected
>So here my question is how did they know that a significance level of .01 was the level they are supposed to compare to?
Did they just choose it because there was no significance level given in the question?

College StatisticsAnswered question
Abram Boyd Abram Boyd 2022-06-26

H 0 : μ = 0 against H A : μ > 0.
Had the following question on my exam today and I'm just wondering if I did this correctly.
From a normally distributed population with mean μ and variance σ 2 a sample has been drawn:
X = ( 0.05 , 4.35 , 0.48 , 0.63 , 1.17 , 2.01 ) ..
a) Test H 0 : μ = 0 against H A : μ > 0 on a 5% significance level under the assumption that σ2 is unknown.
c) Do the same test, given that σ = 1.5..
Solution: I know that H 0 is to be rejected if T c ,, where T is the teststatistic and F t n 1 = 1 α ,, where α is the significance level.
So lets first compute them for a), where we need to use a t-distribution since σ is unknown, so we have that c = F t 5 1 ( 0.95 ) 2.015.. Also, the estimator for σ 2 is
s 2 = 1 n 1 ( k = 1 n X k 2 1 n ( k = 1 n X k ) 2 ) = 1 5 ( 24.961 1 6 ( 41.861 ) ) = 3.405 ,
so, s = 3.405 = 1.845.. We also have that X ¯ = 6.47 / 6 = 1.078..
The teststatistic is
T = X ¯ μ 0 s / n = 1.078 0 1.845 / 6 = 1.078 0.753 = 1.43 c ,
which means that we can not reject H0.
For b) we do an identical approach but wtith s = σ = 1.5 and we use a z-test instead. Here we have that c = Φ 1 ( 0.95 ) = 1.644. The teststatistic in this case is
Z = X ¯ μ 0 σ / n = 1.078 1.5 / 6 = 1.76 c ,
thus we reject H 0 ..
Is this correct?
What my professor does in the solutions is he just makes a 95% confidenceinterval and just arrives to the same conclusions as I've done. Is his method correct without using the teststatistics?

College StatisticsAnswered question
Cory Patrick Cory Patrick 2022-06-25

A company produces millions of 1-pound packages of bacon every week. Company specifications allow for no more than 3 percent of the 1-pound packages to be underweight. To investigate compliance with the specifications, the company’s quality control manager selected a random sample of 1,000 packages produced in one week and found 40 packages, or 4 percent, to be underweight.
Assuming all conditions for inference are met, do the data provide convincing statistical evidence at the significance level of α = 0.05 that more than 3 percent of all the packages produced in one week are underweight?
(A) Yes, because the sample estimate of 0.04 is greater than the company specification of 0.03.
(B) Yes, because the p-value of 0.032 is less than the significance level of 0.05.
(C) Yes, because the p-value of 0.064 is greater than the significance level of 0.05.
(D) No, because the p-value of 0.032 is less than the significance level of 0.05.
(E) No, because the p-value of 0.064 is greater than the significance level of 0.05.
The answer is (B) and I was trying to understand why. My calculation was:
H a : p > 0.03
z = p ^ p 0 p 0 ( 1 p 0 ) n
z = 0.04 0.03 ( 0.03 ) ( 0.07 ) 1000
But I get a ridiculously high number. I'm confused about how to get the p-value in this case.
A two-sided t-test for a population mean is conducted of the null hypothesis H 0 : μ = 100. If a 90 percent t-interval constructed from the same sample data contains the value of 100, which of the following can be concluded about the test at a significance level of α = 0.10?
(A) The p-value is less than 0.10, and H 0 should be rejected.
(B) The p-value is less than 0.10, and H 0 should not be rejected.
(C) The p-value is greater than 0.10, and H 0 should be rejected.
(D) The p-value is greater than 0.10, and H 0 should not be rejected.
(E) There is not enough information given to make a conclusion about the p-value and H 0 .
Here the answer is D, but again I am confused. How can I find the p-value in this case?

Almost all significance test practice problems that you will encounter below help to find solutions to your questions as the answers deal with the same equations that have been used. Start with any significance test example and you will understand that you only have to change variables to determine each value. It's exactly what makes significance test equation so popular as it provides help with more advanced probability concepts. As you look through significance test questions, look for similar patterns as these are where you must start regardless of whether you deal with a complex engineering project or statistical analysis.