Hypotheses construction for significance testing

In summary: Since the problem is talking about a small subset of the population of junior high school students, I don't see how talking about mu (a population parameter) in this context would be correct.
  • #1
Mr Davis 97
1,462
44

Homework Statement


State the null and alternative hypotheses.

The decision to implement changes in the current math program at a junior high will be based on a sample of students' scores on a standardized math exam. If the average is less than the statewide average of 89, all math teachers will have to participate in a workshop to revise the curriculum.

Homework Equations


None.

The Attempt at a Solution


Based on previous examples I've seen, I would write:
##\textrm{H}_{0}: \mu = 89##
##\textrm{H}_{a}: \mu < 89##

However, I have no idea why this would be correct or incorrect. Essentially, I want to know why this would be the correct answer (if it is). Since the problem is talking about a small subset of the population of junior high school students, I don't see how talking about mu (a population parameter) in this context would be correct. If the sample of students at the junior high school scored lower than 89 simply because they're not so smart (i.e. not due to random variation), why would we reject the null and conclude that the mu, which refers to the entire population, is lower based on that one sample? I don't understand.
 
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  • #2
Mr Davis 97 said:
Since the problem is talking about a small subset of the population of junior high school students, I don't see how talking about mu (a population parameter) in this context would be correct.

Applying statistics to real life problems is subjective, so don't expect to find rigorous mathematical justification for using a particular method in real life. The concepts of "reject" and "accept" are not the same as the mathematical concepts of "prove" and "disprove". Hypothesis testing is a procedure not a method of proof. Decisions reached by applying it are not guaranteed to be correct. The justification for using hypothesis testing is empirical. Forms of it have been found useful. There are many different ways to do hypothesis testing. Particular ways of doing it have become traditional in some fields of science.

Applying statistics to textbook exercises is reasonably objective because the author of the textbook expects the problem to be solved in a particular way!

Is this is merely a verbal problem about the concept of "null hypothesis", a problem that gives no specific test scores?

You have to understand the conventions of the textbook. The textbook may want you to say that the null hypothesis is [itex] \mu \ge 89 [/itex] because there would be no justification for revising the curriculum if it was "better than average". On the other hand, to work a problem with given data, you must have a hypothesis that gives enough information to do numerical computations. For example, if you had the students' test scores, then a numerical calculation could be done with a probability distribution that had [itex] \mu = 89 [/itex] , while the information [itex] \mu \ge 89 [/itex] doesn't describe a specific distribution.

You have to determine what convention the textbook uses. Does it insist that the null hypothesis is precisely the logical negation of the "alternative hypothesis"? Or does it say the null hypothesis is the information used to do a numerical calculation? There may be books that aren't consistent in this terminology, in which case you have to figure out what's going on in the current chapter.
 
  • #3
Mr Davis 97 said:

Homework Statement


State the null and alternative hypotheses.

The decision to implement changes in the current math program at a junior high will be based on a sample of students' scores on a standardized math exam. If the average is less than the statewide average of 89, all math teachers will have to participate in a workshop to revise the curriculum.

Homework Equations


None.

The Attempt at a Solution


Based on previous examples I've seen, I would write:
##\textrm{H}_{0}: \mu = 89##
##\textrm{H}_{a}: \mu < 89##

However, I have no idea why this would be correct or incorrect. Essentially, I want to know why this would be the correct answer (if it is). Since the problem is talking about a small subset of the population of junior high school students, I don't see how talking about mu (a population parameter) in this context would be correct. If the sample of students at the junior high school scored lower than 89 simply because they're not so smart (i.e. not due to random variation), why would we reject the null and conclude that the mu, which refers to the entire population, is lower based on that one sample? I don't understand.

Try to think of the effects of "randomness". If the current program had a theoretical average of 89 or better, we would be happy to keep it, but even in that case, the average in a small sample (say in a single junior high) could come out a bit less than 89, just by "chance". Realizing this, you might say that if the sample's average is a bit less than 89 there is not enough justification to warrant the inconvenience, disruption and expense of revising the curriculum. However, if the school's average comes out a lot less than 89, you would be strongly inclined to say the curriculum needs revision. So, in a problem like this you might typically test ##\text{H}_0: \mu = 89## vs. ##\text{H}_1: \mu < 89##, and reject ##\text{H}_0##, hence, accept ##\text{H}_1##---thus taking some possibly expensive revision actions---if the sample mean falls below some critical value ##m < 89##. You would typically like to choose ##m## so that there would not be too great a chance of adjusting the process unneccessarily. That is, even though we might wrongly adjust a perfectly good current system occasionally (just because of random fluctuations), we do not want to do that too often. So, if ##\mu = 89## is true, you want to use a critical value ##m < 89## so that ##P(\text{sample average} \leq m)= p##, for some "small" ##p##, such as ##p =## 0.10 or .05 or .01, for example.
 
  • #4
Ray Vickson said:
Try to think of the effects of "randomness". If the current program had a theoretical average of 89 or better, we would be happy to keep it, but even in that case, the average in a small sample (say in a single junior high) could come out a bit less than 89, just by "chance". Realizing this, you might say that if the sample's average is a bit less than 89 there is not enough justification to warrant the inconvenience, disruption and expense of revising the curriculum. However, if the school's average comes out a lot less than 89, you would be strongly inclined to say the curriculum needs revision. So, in a problem like this you might typically test ##\text{H}_0: \mu = 89## vs. ##\text{H}_1: \mu < 89##, and reject ##\text{H}_0##, hence, accept ##\text{H}_1##---thus taking some possibly expensive revision actions---if the sample mean falls below some critical value ##m < 89##. You would typically like to choose ##m## so that there would not be too great a chance of adjusting the process unneccessarily. That is, even though we might wrongly adjust a perfectly good current system occasionally (just because of random fluctuations), we do not want to do that too often. So, if ##\mu = 89## is true, you want to use a critical value ##m < 89## so that ##P(\text{sample average} \leq m)= p##, for some "small" ##p##, such as ##p =## 0.10 or .05 or .01, for example.

Stephen Tashi said:
Applying statistics to real life problems is subjective, so don't expect to find rigorous mathematical justification for using a particular method in real life. The concepts of "reject" and "accept" are not the same as the mathematical concepts of "prove" and "disprove". Hypothesis testing is a procedure not a method of proof. Decisions reached by applying it are not guaranteed to be correct. The justification for using hypothesis testing is empirical. Forms of it have been found useful. There are many different ways to do hypothesis testing. Particular ways of doing it have become traditional in some fields of science.

Applying statistics to textbook exercises is reasonably objective because the author of the textbook expects the problem to be solved in a particular way!

Is this is merely a verbal problem about the concept of "null hypothesis", a problem that gives no specific test scores?

You have to understand the conventions of the textbook. The textbook may want you to say that the null hypothesis is [itex] \mu \ge 89 [/itex] because there would be no justification for revising the curriculum if it was "better than average". On the other hand, to work a problem with given data, you must have a hypothesis that gives enough information to do numerical computations. For example, if you had the students' test scores, then a numerical calculation could be done with a probability distribution that had [itex] \mu = 89 [/itex] , while the information [itex] \mu \ge 89 [/itex] doesn't describe a specific distribution.

You have to determine what convention the textbook uses. Does it insist that the null hypothesis is precisely the logical negation of the "alternative hypothesis"? Or does it say the null hypothesis is the information used to do a numerical calculation? There may be books that aren't consistent in this terminology, in which case you have to figure out what's going on in the current chapter.

Thank you for the input, it helps. However, my main question still seems to be unanswered. My main question is why in this problem are we basing what the population mean test score is, 89, on whether this specific junior high school in question has a mean below or around 89? Why should we reject the null hypothesis that the mean test score in the state is 89 given that the one specific school scores much lower than 89? Wouldn't this just indicate that the students in the specific school just aren't very smart, rather than indicate that the population mean is lower than 89?
 

1. What is a hypothesis in significance testing?

A hypothesis in significance testing is a statement or prediction about the relationship between variables that is tested using statistical analysis. It is a key part of the scientific method and is used to make predictions and draw conclusions about a population based on a sample of data.

2. How do you construct a hypothesis for significance testing?

To construct a hypothesis for significance testing, you must first identify the research question and the variables involved. Then, you can formulate a null hypothesis, which states that there is no significant difference or relationship between the variables, and an alternative hypothesis, which states that there is a significant difference or relationship between the variables.

3. What is the significance level in hypothesis construction?

The significance level in hypothesis construction is the predetermined threshold for determining whether the results of a statistical analysis are statistically significant. It is typically set at 5% (or 0.05) and represents the probability of obtaining a result at least as extreme as the one observed if the null hypothesis is true.

4. How do you test the significance of a hypothesis?

To test the significance of a hypothesis, you can use statistical tests such as t-tests, ANOVA, or chi-square tests. These tests calculate a p-value, which is the probability of obtaining a result at least as extreme as the one observed if the null hypothesis is true. If the p-value is lower than the significance level, the results are considered statistically significant and the null hypothesis is rejected.

5. What are the potential limitations of hypothesis construction for significance testing?

One limitation of hypothesis construction for significance testing is that it can only determine whether there is a statistically significant difference or relationship between variables, but it cannot prove causation. Additionally, the results may be influenced by sample size, the chosen significance level, and potential confounding variables. It is important to consider these limitations when interpreting the results of a significance test.

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