Confidence Intervals & Hypothesis Testing: MATH 130 Example, Study notes of Statistics

Download Confidence Intervals & Hypothesis Testing: MATH 130 Example and more Study notes Statistics in PDF only on Docsity! Millersville University Department of Mathematics MATH 130, Review of Confidence Intervals and Hypothesis Testing April 7, 2008 1. Five thousand college graduates were asked how many years were required to earn their bachelors degrees. The sample mean was 5.2 years with a standard deviation of 1.7 years. Find the 98% confidence interval for the mean graduation time for all college students. This is a large sample (n = 5000 > 30) with mean x = 5.2 and s = 1.7. Note that the population standard deviation is not known, so we will use the Student’s t-distribution for the distribution of the sample means. At the 98% confidence level the value of α = 0.02 and thus α/2 = 0.01. The critical value is tα/2 = 2.326 according to Table V. Thus the margin of error in this estimate of the mean is E = tα/2 s√ n = (2.326) ( 1.7√ 5000 ) ≈ 0.1 Note that we are rounding the margin of error to one decimal place just like the mean and standard deviation. Therefore the 98% confidence interval estimate of the mean number of years of college before graduation for all college students is (x − E, x + E) = (5.2 − 0.1, 5.2 + 0.1) = (5.0, 5.3) 2. If the standard deviation in the price of a new car is $14, 138 how many car prices must be sampled if the error of the sample mean should be no greater than $2000 at the 90% confidence level? According to the problem statement, σ = 14, 138. The margin of error for the sample should be E = 2000. At the 90% confidence level the value of α = 0.10 and thus α/2 = 0.05. The critical value is zα/2 = 1.645 according to Table V. Thus the sample size needed for this estimate is n = ( zα/2 E ) 2 σ2 = ( 1.645 2000 )2 (14138)2 = 135.22 ≈ 136 Notice that we must always round sample sizes up to the next whole number. 3. In researching automobile maintenance costs 15 service stations were asked the price of an oil change. The mean was $17.56 with a standard deviation of $3.41. Construct the 99% confidence interval for the mean cost of an oil change. This is a small sample (n = 15 ≤ 30) with mean x = 17.56 and s = 3.41. The population standard deviation is unknown so we will use the Student’s t-distribution for the distribution of the sample means. The number of degrees of freedom for this sample is df = n− 1 = 15− 1 = 14. At the 99% confidence level the value of α = 0.01. Thus α/2 = 0.005 and tα/2 = 2.977 according to Table V. Thus the margin of error in this estimate of the mean is E = tα/2 s√ n = (2.977) ( 3.41√ 15 ) ≈ 2.62 Note that we are rounding the margin of error to two decimal places just like the mean and standard deviation. Therefore the 99% confidence interval estimate of the average price of an oil change is (x − E, x + E) = (17.56 − 2.62, 17.56 + 2.62) = (14.94, 20.18) 4. An engineer claims that a new fuel injection design increases the mean mileage on a car above its current 31 mpg. (a) Express the claim in symbolic form. If µ represents the mean mileage of cars with the new fuel injection design then the claim above can be expressed in the symbolic form: µ > 31. (b) Identify the null hypothesis. Since the null hypothesis always contains a condition of equality then the null hypothesis can be expressed symbolically as, H0 : µ = 31. (c) Identify the alternative hypothesis. The alternative hypothesis does not contain a condition of equality, thus the alter- native hypothesis is the original claim in this case. It can be expressed symbolically as, H1 : µ > 31. (d) Identify the test as being left-tailed, right-tailed, or two-tailed. The alternative hypothesis can be used to determine whether a test is left-tailed, right-tailed, or two-tailed. The inequality symbol always “points” in the direction of the tail of the test. In this example the inequality symbol is “greater than” (>) and therefore this is a right-tailed test. (e) Assuming the conclusion is to reject the null hypothesis, state the conclusion in nontechnical terms. Since the original claim is the alternative hypothesis and the null hypothesis is rejected, the conclusion can be stated as, “The sample data support the claim that the mean mileage of cars with the new fuel injection design is greater than 31 mpg.” (f) Assuming the conclusion is failure to reject the null hypothesis, state the conclu- sion in nontechnical terms. Since the original claim is the alternative hypothesis and the null hypothesis is not rejected, the conclusion can be stated as,