Hypothesis Test for a Single Proportion using StatCrunch in MyMathlab MyStatlab | Summary and Q&A

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May 17, 2018
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The Math Sorcerer
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Hypothesis Test for a Single Proportion using StatCrunch in MyMathlab MyStatlab

TL;DR

Test the claim that more than 20% of subjects developed nausea after taking a pain treatment using a significance level of 0.05.

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Key Insights

  • 🤢 The problem involves a hypothesis test for a proportion in the context of subjects developing nausea after a pain treatment.
  • ❓ The null hypothesis assumes a proportion of 20% or less, while the alternative hypothesis suggests a proportion greater than 20%.
  • 🏆 StatCrunch is used to calculate the test statistic, which is compared to the significance level to determine the conclusion.
  • 🏆 The p-value for this hypothesis test is 0.246, indicating that the observed data is not significantly different from what would be expected under the null hypothesis.

Transcript

in this problem we have to do a hypothesis test for a proportion suppose 234 subjects are treated with a drug that is used to treat pain and 51 of them developed nausea use a point O 5 significance level to test the claim that more than 20% of them developed nausea okay the first step is to identify the null and alternative hypotheses so the last s... Read More

Questions & Answers

Q: What is the null and alternative hypothesis in this hypothesis test?

The null hypothesis states that the proportion of subjects with nausea is 20% or less, while the alternative hypothesis claims that the proportion is greater than 20%.

Q: How do you compute the test statistic using StatCrunch?

In StatCrunch, under proportion stats, select one sample with summary. Enter the number of successes (those who developed nausea) and the total number of subjects. Set the null and alternative hypotheses to match the problem statement, and click compute to obtain the test statistic.

Q: What is the p-value for this hypothesis test?

The p-value is 0.246, rounded to three decimal places. It represents the probability of obtaining the observed data or more extreme data if the null hypothesis is true.

Q: What is the conclusion for this hypothesis test?

Since the p-value (0.246) is greater than the significance level (0.05), we fail to reject the null hypothesis. There is not sufficient evidence to support the claim that more than 20% of users develop nausea.

Summary & Key Takeaways

  • The problem requires a hypothesis test for a proportion regarding the development of nausea after using a pain treatment.

  • The null hypothesis is that the proportion is 20% or less, while the alternative hypothesis states that the proportion is greater than 20%.

  • StatCrunch is used to compute the test statistic and p-value, which is compared to the significance level to determine the conclusion.

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