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Testing a Claim about a Proportion Assumptions 1.The sample was a simple random sample 2.The conditions for a binomial distribution are satisfied 3.Both np ≥ 5 and nq ≥ 5 Test Statistic:

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Step-by-step through an example: Suppose 5% of people normally get the flu during flu season. To test a new flu vaccine, the researchers give the vaccine to 1000 people. Of these, 43 got the flu. Test the claim that the infection rate for those taking the vaccine was less than 5%, using a 0.01 significance level.

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Step 1. Write the original claim in symbolic form. Step 2. Write the opposite of the original claim Step 3. Of the two statements, choose the one without equality to be the alternative hypothesis. The null hypothesis will be that p is equal to the fixed value (in this case 0.05)

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Step 4. Select the significance level. In this case, we were told to use α = 0.01 Step 5. Identify the relevant statistic (characteristic of the sample). Since we are testing a claim about the population proportion p, the statistic is relevant. The distribution of sample proportions is approximately normal.

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Traditional Method Step 6. Determine the value of the test statistic, under the assumption of the null hypothesis (assuming p = 0.05) This is a left-tailed test, so the critical region is in an area of α = 0.01 in the left tail. Using the table, we find the critical value to be z = -2.33.

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Traditional Method Step 7. Since the test statistic does not fall within the critical region, we fail to reject the null hypothesis. Step 8. We conclude that there is not sufficient evidence to support the claim that those who took the vaccine had an infection rate less than 5%.

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P-Value Method Step 6. We find the probability of a value at least as extreme as our sample result. Since this is a left-tailed test, we want to find the area to the left of our test statistic (see pg 378) P(z < -1.016) = 0.1539. This is our P-value. Step 7. Because the P-value is not less than the significant level of 0.01, we fail to reject the null hypothesis.

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Relationship to Confidence Intervals (a.k.a. Confidence Interval Method) We will find the 99% confidence interval for the population proportion p based on our sample statistic of Since the value of 0.05 is included in the confidence interval, we do not have sufficient evidence to suggest that p is less than 0.05.

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Testing a Claim about a Mean Assumptions 1.The sample was a simple random sample 2.Either or both of these conditions is satisfied: The population is normally distributed, or n > 30. Test Statistic: if σ is knownif σ is unknown

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An Example IQ tests are designed to have a mean of 100. We want to test the claim that people hit over the head before taking an IQ test will score differently. To test this, we hit 61 people over the head and give them an IQ test. The samples returns a mean of 95, with a standard deviation of 10.

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Step 1. Write the original claim in symbolic form. Step 2. Write the opposite of the original claim Step 3. Of the two statements, choose the one without equality to be the alternative hypothesis. The null hypothesis will be that μ is equal to the fixed value (in this case 100)

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Step 4. Select the significance level. Lets use α = 0.05 Step 5. Identify the relevant statistic (characteristic of the sample). Since we are testing a claim about the population proportion μ, the statistic is relevant. The distribution of sample means has a t-distribution, since sigma is unknown.

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Traditional Method Step 6. Determine the value of the test statistic, under the assumption of the null hypothesis (assuming μ = 100). Since we don’t know σ, we will use the t test statistic. This is a two-tailed test, so the critical region is in an area of α = 0.05 in the two tails. Using the critical t-values table (with 61-1=60 degrees of freedom), we find the critical values to be t = -2.000 and t = 2.000.

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Traditional Method Step 7. Since the test statistic does fall within the critical region, we reject the null hypothesis. Step 8. We conclude that the sample data support the claim that the mean IQ score for people hit over the head before taking an IQ test is different than 100.

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P-Value Method Step 6. We find the probability of a value at least as extreme as our sample result. Since this is a two-tailed test and our sample fell to the left of the assumed mean, we find the area to the left of our test statistic and double it (see pg 378) P(t < -3.905) = 0.0001 (You do not have a table for this – I calculated this using my calculator) Double this to 0.0002. This is our P-value. Step 7. Because the P-value is less than the significant level of 0.05, we reject the null hypothesis. Note: Since you do not have any good way to calculate P-values for t distributions, I will not ask you to do the P-value method for tests of means with sigma unknown.

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Homework 7.3: 7, 9, 13, 15 7.4: 9 7.5: 9, 11, 15, 17

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Created by Erin Hodgess, Houston, Texas Section 7-5 Testing a Claim About a Mean: Not Known.

Created by Erin Hodgess, Houston, Texas Section 7-5 Testing a Claim About a Mean: Not Known.

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