# Help!! | Mathematics homework help

### LESSON

#### Lesson Objectives:

 Student will write null and alternative hypotheses. Student will find critical values for testing a mean or proportion against the population mean or proportion using the appropriate test, based on the sample size. Student will find critical values for testing the difference between two means or two proportions using the appropriate test, based on the sample size.

#### What is a hypothesis?

In science, you may have learned that the hypothesis is an educated guess.  In statistics, the same definition carries over but has some different applications.  A statistical study is similar to the scientific method.  From science you have learned that the scientific method includes the following steps:  1) Ask a question 2) Do background research 3) Construct a hypothesis 4) Test your hypothesis by doing an experiment 5) Analyze the data and draw a conclusion 6) Communicate the results.  In statistics, there are two hypotheses that need to be formed once you have defined the problem and completed background research.  One is called the “null” hypothesis and the other is called the “alternative” hypothesis.  Once the study is conducted, we can reject or fail to reject either of the hypothesis based on the results of the study.

The Null Hypothesis

The null hypothesis is composed of the fact that there is no effect of the treatment on the subjects in the study.  For example if we were trying to investigate the relationship between two variables our null hypothesis may state that “there is no relationship between the two variables” or if we are trying to see if a new drug has an effect on weight gain the null hypothesis may state that “the drug has no effect on the weight gain of the subjects”.  The null hypothesis is the one that we will fail to reject (accept) unless the data provides convincing evidence that it is false.

The Alternative Hypothesis

The alternative hypothesis may be referred to as the opposite of the null hypothesis.  For example, if the null hypothesis states that there is no relationship between two variables, then the alternative hypothesis should state that “there is a relationship between the two variables that can be measured.  If the null hypothesis states that there is no effect on the subject then the alternative hypothesis should state that “there is an effect on the subject”.  We will fail to reject (accept) the alternative hypothesis if and only if the data provides convincing evidence that it is true.

#### Practice Writing Null and Alternative Hypotheses

The hypotheses can be written out in words or we may use mathematical symbols to express the hypothesis.  Here a few examples of how to write the null and alternative hypothesis.  The most common symbol for the null hypothesis is H0 and the most common symbol for the alternative hypothesis is H1.

Let’s Practice:

Case I: An agriculturist is doing a study to determine if a fertilizer has any effect on the average height of 100 apple trees.  He knows that the average height of unfertilized apple trees is 10ft.  The average height of the 100 apple trees that were treated with fertilizer is 10.8 feet with a standard deviation of .5 ft.

1)  Do you think that the fertilizer has an effect on the height of the apple trees?
H0 = The fertilizer has no effect on the height of the apple trees                         Sample mean = Population mean
H1 = The fertilizer does have an effect on the height of the apple trees.             Sample mean
≠ Population mean

2)
Does the fertilizer make the apple trees taller?
H= The fertilizer does not make the apple trees taller.                                     Sample mean = Population mean
H1 = The fertilizer does make the apple trees taller.                                           Sample mean > Population mean

Remember that when writing and testing hypotheses it is very important that you consider the question that want to answer with your study because this fact helps to shape the correct hypotheses.

#### Choosing the Appropriate Test

Choosing the appropriate test for any statistical research is very important to obtaining the most accurate results.  We discussed in a previous lesson when to use a z-score or z-test and when to use a t-score or a t-test.  Recall that we use a z-test when the sample size is fairly large (greater than 100) and a t-test when the sample size is small (less than 100).  Another question that we need to consider is when to use a two- tailed test or a one- tailed test.  This choice is made based upon the statement of the alternative hypothesis.  We would use a two-tailed test if our alternative hypothesis is the exact opposite of the null hypothesis.  We would use a one-tailed test if our alternative hypothesis suggests a certain direction for the results.

If  H0 = There is no effect on the variable , then H1 = There is an effect on the variable.

In this case we would use a two tailed test.

If H0 = The mean sample mean is equal to the population mean , then H1 = The sample mean is greater than the population mean.

In this case we would use the one-tailed test since we are concerned with only the right side of the probability distribution where the values are that are greater than the mean.

Click here for a video explaining when to use a one-tailed test and when to use a two tailed test.

#### Critical Values & Testing a Hypothesis

Testing a hypothesis includes finding the appropriate z or t score and finding critical values to compare it to the probability of an event and its position on the normal distribution curve given a confidence interval or p-value.  A critical value is the value of the dependent variable at a given point of a function that helps us to decide to reject or fail to reject the null hypothesis.  If the test statistic is this number or more then we will reject the null hypothesis but if the test statistic is less than this number then we will fail to reject or accept the null hypothesis.

Consider these examples:

Testing the mean against the population mean

An agriculturist is doing a study to determine if a fertilizer has any effect on the average height of 100 apple trees.  He knows that the average height of unfertilized apple trees is 10ft.  The average height of the 100 apple trees that were treated with fertilizer is 10.1 feet with a standard deviation of 0.5 ft.  Do you think that the fertilizer has an effect on the height of the apple trees?

Step 1:  Write the null and alternative hypotheses
H0 = The fertilizer has no effect on the height of the apple trees                         Sample mean = Population mean
H1 = The fertilizer does have an effect on the height of the apple trees.             Sample mean
≠ Population mean

Step 2:  We will assume that the null hypothesis is true and find the z-score (since we have a large sample).

z = 10 – 10.1/(.5/√100) = -2

Step 3:  Find the critical value by applying the subtracting (since the z-score is negative) the product of the z-score and the standard deviation from the population mean.

10 – 2(.5) = 9 so the critical value is 9

Since the mean of the sample is greater than this critical value we must reject the null hypothesis.

Testing a proportion against the population proportion

A company found that in a sample of 100 of its products that 25 were defective after retraining the employees.  If there is an overall 60% chance that the company will produce a defective product, did the training help employees minimize the number of defective products?

Step 1:
Write the null and alternative hypothesis.

H0 = p = 60% or .6        The training had no effect on the proportion of defective products
H= p < 60% or .6        The training helped to decrease the proportion of defective products.

Remember that these types of problems are binomial experiments so we have to be sure that we can use the normal approximation by confirming that (n)(p) and (n)(q) are both greater than 5.  So (100)(.6) = 60 and (100)(.4) = 40, we may proceed.

Step 2:

Assume that the null hypothesis is true and find the standard deviation ( standard deviation = √(n)(p(q)) so that we can find the z-score.

standard deviation = 4.33

z = .25 – .6/ 4.33 = -.08

Step 3:  Find the critical value by subtracting (since the z-score is negative) the product of the z-score and the standard deviation from the mean.

.6 – .3464 = .2536 or 25.36 %  since our test statistic is .6 or 60% and it is more than the critical value then we must reject the null hypothesis and conclude that training did decrease the number of defective products.

#### Testing the Difference of Means and Difference of Proportions

The difference of means or difference of proportions is used to find out whether there is a significance between the controlled group and the uncontrolled group.  We can also use critical values to determine the significance for the difference of means as well as the difference of proportions.

Difference of Means:

A teacher gave 100 students a study guide in preparation for a major test.  The average score for these students was 88 with standard deviation of 2.  She did not give study guides to another group of 100 students and the average score for these students was 80 with a standard deviation of 3.  Determine whether the study guide had an effect on student scores.

Step 1:  Find the difference in the sample mean scores.  88-80 = 8

Step 2:  Write the null and alternative hypotheses.
H0 = The study guide has no effect on student test scores or population mean1 = population mean2 or population mean1 – population mean2 = 0
H1 = The study guide has an effect on student test scores or population mean1 ≠ population mean2 or population mean1 – population mean2 ≠ 0

Step 3:  Calculate the standard deviation for the difference of the sample means.
To do this we need to find the variance (square the standard deviation for each sample) and divide it by n for each sample then add the two values together and take the square root.

√[(22 )/100 + (32 )/100] = √4/100 + 3/100 = √7/100 = √.07 = .26

Step 4:  Assume that the null hypothesis is true and find the z-score using the difference of the means.
z = 0-8/.26 = -30.76  find the critical value by adding (since the z-score is positive) the product of the z-score and the standard deviation to the test statistic.

0 – 8 = -8 This is the critical value and since the test statistic 0 is must than this value we must reject the null hypothesis and conclude that the study guide has an effect on student test scores.

NEXT TEACHER OFFICE HOURS ARE:

#### Assignment:

For questions 1-5, write the null and alternative hypothesis.

1.  Does the water temperature have an effect on the number of people in the pool?

2.  Does the weather have an effect on the number of people at the beach?

3.  A fitness center is running a discounted membership fee.  Did the discount increase the membership sales? Write your hypotheses mathematically.

4.  A medical researcher gave 100 patients a new drug to see if it reduces their blood pressure?  Did the new drug reduce the patients’ blood pressure?  Write your hypothesis mathematically.

5.  Some students took a conflict resolution class?  Did this class help to reduce the number of conflicts that the students were involved in?  Write your hypothesis mathematically.

For questions 6-10, find the critical value.

6.  There is an annual hot dog eating contest in Plattsburg, MS and the average number of hot dogs eaten by one person is 36 with a standard deviation of 6.  Find the critical value for a person who can eat more than 2 deviations above the mean.

7.  Female high school seniors at a particular school have an average height of 65 inches with a standard deviation of 5 inches.  Find the critical value for a female high senior who is less than 1 deviation from the mean.

8.  The average weight of a newborn at a particular hospital is 96 ounces with a standard deviation of 3 ounces.  Find the critical value for a newborn who is 2 standard deviations below the average weight.

9.  A company found that 27 out of 150 of its products were defective after retraining its employees.  If the company normally has a 40% defective product rate, find the critical value to determine if retraining the employees helped to minimize the number of defective products.

10.  A teacher gives 200 students a study guide for a test and the average score was 90 with a standard deviation of 6.  She did not give the other 200 students a study guide and their average score was 70 with a standard deviation of 8.  Find the critical value to determine whether or not the study guide helped students to increase their test score.

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