When conducting a hypothesis test with a binomial distribution (sometimes called a Binomial Test), there are three ways to calculate the P-value (with additional variations possible). The only exact calculation is to use the binomial probability distribution. The other methods are approximations using the standardized normal distribution (when certain criteria have been achieved). Of these two methods, one can use the sample counts or one can use the sample proportions. Furthermore, it is possible in both of these approximating cases to apply a continuity correction to account for the use of a continuous distribution to approximate a discrete distribution. This problem introduces the method to obtain an exact P-value using the binomial distribution. Also, this method uses Excel to obtain the answers. For this demonstration problem, we will test a hypothesis that a population proportion has increased since the last time it was measured. Previously, the population proportion was measured at 20%. For the current analysis, a sample of n=150n=150 randomly chosen subjects was obtained, and 37 of those demonstrated the observation of interest (i.e., a success). To start, we clearly construct the hypotheses for this problem. Because the researcher suggested the proportion has increased, this would suggest a one-tailed test (as can be seen in the choice of HaHa): Ho:p=0.2Ho:p=0.2 Ha:p>0.2Ha:p>0.2 Using the binomial distribution, the test statistic from the sample would simply be the sample count (the number of successful observations): k=37k=37 The P-value for this scenario would be observing this count or one more extreme. With the alternative hypothesis suggesting that values at or below 20% would be unsurprising, this would suggest that the observed count or larger would constitute the potentially extreme responses. Thus, the P-value would be: P(X≥37∣p=0.2,n=150)P(X≥37∣p=0.2,n=150) Using Excel, this value can be calculated exactly:=1-BINOMDIST(36,150,20%,TRUE)which should return a P-value of 0.0945. Thus, with a traditional significance level of either α=0.05α=0.05 or α=0.01α=0.01, this P-value would result in failing to reject the null hypothesis. Thus, there is not enough sample evidence to support the claim that the population proportion has increased. Note: For a two-tailed hypothesis test, the calculation can become a bit tricky. In this case, it is necessary to take the counts at and above the sample count, but it is also necessary to determine comparable counts below the hypothesized population count. This would be obtained using the formulan⋅p−(k−n⋅p)n⋅p-(k-n⋅p)This value is the count below the hypothesized mean count by the same distance as the sample count was above the mean count. For this demonstration example, this value would be 23. To obtain the P-value from Excel, you would use the following formula:=BINOMDIST(23,150,20%,TRUE)+1-BINOMDIST(36,150,20%,TRUE)Though, it is recommended that the two-tailed test only be used with one of the approximation methods as the calculations are less cumbersome. Exercise Problem In 1977, only 10% of the students in the city school district were classified as being learning disabled. A school psychologist suspects that the proportion of learning-disabled children has increased dramatically over the years. To demonstrate this point, a random sample of n=200n=200 students is selected. In this sample there are 27 students who have been identified as learning-disabled. You will use this information to determine if the sample indicates a change in the proportion of learning-disabled students at a 0.02 level of significance. With these hypotheses, the p-value for this test is (assuming HoHo is true) the probability of observing... at most 27 learning-disabled students at least 27 learning-disabled students more than 27 learning-disabled students at least 20 learning-disabled students This P-value (and test statistic) leads to a decision to... reject the null accept the null fail to reject the null reject the alternative Incorrect

When conducting a hypothesis test with a binomial distribution (sometimes called a Binomial Test), there are three ways to calculate the P-value (with additional variations possible). The only exact calculation is to use the binomial probability distribution. The other methods are approximations using the standardized normal distribution (when certain criteria have been achieved). Of these two methods, one can use the sample counts or one can use the sample proportions. Furthermore, it is possible in both of these approximating cases to apply a continuity correction to account for the use of a continuous distribution to approximate a discrete distribution. This problem introduces the method to obtain an exact P-value using the binomial distribution. Also, this method uses Excel to obtain the answers. For this demonstration problem, we will test a hypothesis that a population proportion has increased since the last time it was measured. Previously, the population proportion was measured at 20%. For the current analysis, a sample of n=150n=150 randomly chosen subjects was obtained, and 37 of those demonstrated the observation of interest (i.e., a success). To start, we clearly construct the hypotheses for this problem. Because the researcher suggested the proportion has increased, this would suggest a one-tailed test (as can be seen in the choice of HaHa): Ho:p=0.2Ho:p=0.2 Ha:p>0.2Ha:p>0.2 Using the binomial distribution, the test statistic from the sample would simply be the sample count (the number of successful observations): k=37k=37 The P-value for this scenario would be observing this count or one more extreme. With the alternative hypothesis suggesting that values at or below 20% would be unsurprising, this would suggest that the observed count or larger would constitute the potentially extreme responses. Thus, the P-value would be: P(X≥37∣p=0.2,n=150)P(X≥37∣p=0.2,n=150) Using Excel, this value can be calculated exactly:=1-BINOMDIST(36,150,20%,TRUE)which should return a P-value of 0.0945. Thus, with a traditional significance level of either α=0.05α=0.05 or α=0.01α=0.01, this P-value would result in failing to reject the null hypothesis. Thus, there is not enough sample evidence to support the claim that the population proportion has increased. Note: For a two-tailed hypothesis test, the calculation can become a bit tricky. In this case, it is necessary to take the counts at and above the sample count, but it is also necessary to determine comparable counts below the hypothesized population count. This would be obtained using the formulan⋅p−(k−n⋅p)n⋅p-(k-n⋅p)This value is the count below the hypothesized mean count by the same distance as the sample count was above the mean count. For this demonstration example, this value would be 23. To obtain the P-value from Excel, you would use the following formula:=BINOMDIST(23,150,20%,TRUE)+1-BINOMDIST(36,150,20%,TRUE)Though, it is recommended that the two-tailed test only be used with one of the approximation methods as the calculations are less cumbersome. Exercise Problem In 1977, only 10% of the students in the city school district were classified as being learning disabled. A school psychologist suspects that the proportion of learning-disabled children has increased dramatically over the years. To demonstrate this point, a random sample of n=200n=200 students is selected. In this sample there are 27 students who have been identified as learning-disabled. You will use this information to determine if the sample indicates a change in the proportion of learning-disabled students at a 0.02 level of significance. With these hypotheses, the p-value for this test is (assuming HoHo is true) the probability of observing... at most 27 learning-disabled students at least 27 learning-disabled students more than 27 learning-disabled students at least 20 learning-disabled students This P-value (and test statistic) leads to a decision to... reject the null accept the null fail to reject the null reject the alternative Incorrect

MATLAB: An Introduction with Applications

6th Edition

ISBN:9781119256830

Author:Amos Gilat

Publisher:Amos Gilat

Chapter1: Starting With Matlab

Section: Chapter Questions

Problem 1P

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When conducting a hypothesis test with a binomial distribution (sometimes called a Binomial Test), there are three ways to calculate the P-value (with additional variations possible). The only exact calculation is to use the binomial probability distribution. The other methods are approximations using the standardized normal distribution (when certain criteria have been achieved). Of these two methods, one can use the sample counts or one can use the sample proportions. Furthermore, it is possible in both of these approximating cases to apply a continuity correction to account for the use of a continuous distribution to approximate a discrete distribution.

This problem introduces the method to obtain an exact P-value using the binomial distribution. Also, this method uses Excel to obtain the answers.

For this demonstration problem, we will test a hypothesis that a population proportion has increased since the last time it was measured. Previously, the population proportion was measured at 20%. For the current analysis, a sample of n=150n=150 randomly chosen subjects was obtained, and 37 of those demonstrated the observation of interest (i.e., a success).

To start, we clearly construct the hypotheses for this problem. Because the researcher suggested the proportion has increased, this would suggest a one-tailed test (as can be seen in the choice of HaHa):
      Ho:p=0.2Ho:p=0.2
      Ha:p>0.2Ha:p>0.2

Using the binomial distribution, the test statistic from the sample would simply be the sample count (the number of successful observations):
      k=37k=37

The P-value for this scenario would be observing this count or one more extreme. With the alternative hypothesis suggesting that values at or below 20% would be unsurprising, this would suggest that the observed count or larger would constitute the potentially extreme responses. Thus, the P-value would be:
      P(X≥37∣p=0.2,n=150)P(X≥37∣p=0.2,n=150)

Using Excel, this value can be calculated exactly:=1-BINOMDIST(36,150,20%,TRUE)which should return a P-value of 0.0945. Thus, with a traditional significance level of either α=0.05α=0.05 or α=0.01α=0.01, this P-value would result in failing to reject the null hypothesis. Thus, there is not enough sample evidence to support the claim that the population proportion has increased.

Note: For a two-tailed hypothesis test, the calculation can become a bit tricky. In this case, it is necessary to take the counts at and above the sample count, but it is also necessary to determine comparable counts below the hypothesized population count. This would be obtained using the formulan⋅p−(k−n⋅p)n⋅p-(k-n⋅p)This value is the count below the hypothesized mean count by the same distance as the sample count was above the mean count. For this demonstration example, this value would be 23. To obtain the P-value from Excel, you would use the following formula:=BINOMDIST(23,150,20%,TRUE)+1-BINOMDIST(36,150,20%,TRUE)Though, it is recommended that the two-tailed test only be used with one of the approximation methods as the calculations are less cumbersome.

Exercise Problem
In 1977, only 10% of the students in the city school district were classified as being learning disabled. A school psychologist suspects that the proportion of learning-disabled children has increased dramatically over the years. To demonstrate this point, a random sample of n=200n=200 students is selected. In this sample there are 27 students who have been identified as learning-disabled. You will use this information to determine if the sample indicates a change in the proportion of learning-disabled students at a 0.02 level of significance.
With these hypotheses, the p-value for this test is (assuming HoHo is true) the probability of observing...
- at most 27 learning-disabled students
- at least 27 learning-disabled students
- more than 27 learning-disabled students
- at least 20 learning-disabled students
- This P-value (and test statistic) leads to a decision to...
  - reject the null
  - accept the null
  - fail to reject the null
  - reject the alternative
  Incorrect
  As such, the final conclusion is that...
  - There is sufficient evidence to warrant rejection of the claim that the proportion of learning-disabled students has increased.
  - There is not sufficient evidence to warrant rejection of the claim that the proportion of learning-disabled students has increased

Features Features Normal distribution is characterized by two parameters, mean (µ) and standard deviation (σ). When graphed, the mean represents the center of the bell curve and the graph is perfectly symmetric about the center. The mean, median, and mode are all equal for a normal distribution. The standard deviation measures the data's spread from the center. The higher the standard deviation, the more the data is spread out and the flatter the bell curve looks. Variance is another commonly used measure of the spread of the distribution and is equal to the square of the standard deviation.

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