Montarello and Martin (2005) found that 5th grade students completed more math problems correctly when simple problems were mixed with their regular math assignments. To further explore this phenomenon, suppose that a researcher selects a standardized math test that produces a normal distribution of μ = 100 and a standard deviation of 18. The researcher modifies the test by inserting a set of very easy problems among the standardized questions and gives the modified test to a sample of 36 students. If the average test score for the sample is M = 104, is this result sufficient to conclude inserting the easy questions improves student performance? Use a one-tailed test with α = 0.05. Also name the independent and dependent variables for this experiment

in a test of a lifelong brand of tires , a survey revealed that the mean life was 35,000 mikes and the standard deviation was 7083 mikes.  if the shape of the population distribution is known to be normal, approximately what percentage of the tires lasts less than 20834 miles ?

Use z scores to compare the given values. The tallest living man at one time had a height of 238 cm. The shortest living man at that time had a height of 142.4 cm. Heights of men at that time had a mean of 175.45 cm and a standard deviation of 5.59 cm. Which of these two men had the height that was more extreme? ... Since the z score for the tallest man is z = 0 and the z score for the shortest man is z = the man had the height that was Im- more extreme. (Round to two decimal places.) shortest tallest

An English teacher has been teaching a sixth grade composition class for many years.  He has the feeling that over the past several years, the writing ability of students has changed.  A national test of proficiency in composition was administered 5 years ago.  The resulting distribution of scores was normally shaped, had a mean of 85 and a standard deviation of 10.9.  In order to test his feeling, he gives his present class of 43 students the same proficiency test.  The resulting mean is 80 and the standard deviation is 8.7.     a. What is the alternative hypothesis?   b. What is the null hypothesis?   c. What is your conclusion, using á = 0.052 tail?   d. What type error may you be making because of your conclusion in part c?

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In a study of birth order and intelligence, IQ tests were given to 18- and 19-year-old men to estimate the size of the difference, if any, between the mean IQs of firstborn sons and secondborn sons. The following data for 10 firstborn sons and 10 secondborn sons are consistent with the means and standard deviations reported in the article. It is reasonable to assume that the samples come from populations that are approximately normal. Can you conclude that the mean IQ of firstborn sons is greater than the mean IQ of secondborn sons? Let μ1 denote the mean IQ of firstborn sons and μ2 denote the mean IQ of secondborn sons. Use the α = 0.01 level and the P-value method with the table. Firstborn 128 101 128 112 121 105 122 98 106 108 Secondborn 121 125 110 107 114 93 80 94 91 83 Part(a): State the appropriate null and alternate hypotheses. H0:   H1: This is a _____…

A hiring company claims that the variance of the annual salaries for managers is greater in Florida than Georgia. You select a sample of 28 managers in Florida and find the standard deviation to be $11,000. You select a sample of 24 managers in Georgia and find the standard deviation to be $9,200. At α=0.05, can you support the hiring company's claim?

A researcher decides to measure anxiety in group of bullies and a group of bystanders using a 23-item, 3 point anxiety scale.  Assume scores on the anxiety scales are normally distributed and the variance among the group of bullies and bystanders are the same. A group of 30 bullies scores an average of 21.5 with a sample standard deviation of 10 on the anxiety scale.  A group of 27 bystanders scored an average of 25.8 with a sample standard deviation of 8 on the anxiety scale. You do not have any presupposed assumptions whether bullies or bystanders will be more anxious so you formulate the null and alternative hypothesis based on that.