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At a local store, 65 female employees were randomly selected and it was found that their...
At a college, 61 female students were randomly selected and it was found that their monthly...
At a college, 61 female students were randomly selected and it was found that their monthly income had a standard deviation of $230.85. For 121 male students, the standard deviation was $320.53. Test the claim that variance of monthly incomes is higher for male students than it is foi female students. Use a = 0.01.
65 female soccer players and 65 male soccer players were randomly selected from the state of...
65 female soccer players and 65 male soccer players were randomly selected from the state of New York. Their body temperatures were taken and were recorded. The mean of the male body temperatures was 98.105 degrees Farenheit and their standard deviation was .699. The mean of the female body temperatures was 98.394 degrees Farenheit and their standard deviation was .743. Is their a significant difference between the means of the two populations? Let alpha be .05. What are the hypotheses...
how do you get the t value in this problem i keep getting the wrong answer....
how do you get the t value in this problem i keep getting the wrong answer. can you show me steps? my answer is not the same as yours, At a local college 53 female students were randomly selected and it was found that their mean monthly income was $612 with a standard deviation of $121.50. 56 male students were also randomly selected and their mean monthly income was found to be $654 with a standard deviationof $168.70. Construct a...
The mean score on a science assessment test for 49 randomly selected male high school students...
The mean score on a science assessment test for 49 randomly selected male high school students was 153. Assume the population standard deviation is 36. The mean score on the same test for 50 randomly selected female high school students was 147. Assume the population standard deviation is 34. At a= 0.05, can you support the claim that the mean score on the science assessment test for male high school students is greater than the mean score for female high...
A store owner surveyed 25 randomly selected customers and found the ages shown (in years). The...
A store owner surveyed 25 randomly selected customers and found the ages shown (in years). The mean is 27.8 and the standard deviation is 8.72. The owner wants to know if the mean age of all customers is 32 years old. Use the given information to complete parts a through f 30 19 23 26 23 D 31 30 23 11 50 30 20 29 18 24 29 3740 46 35 26 23 22 27 23 a) What is the...
A store owner surveyed 25 randomly selected customers and found the ages shown (in years). The...
A store owner surveyed 25 randomly selected customers and found the ages shown (in years). The mean is 32.36 and the standard deviation is 11.79. The owner wants to know f the mean age of all customers is 36 years old. Use the given information to complete parts a through . 14 12 18 22 16 42 48 44 28 40 41 3743 44 38 16 20 39 48 31 30 34 21 47 36 a) What is the null...
The California Health Survey involved 51,048 randomly selected adults in California. The author selected 100 of...
The California Health Survey involved 51,048 randomly selected adults in California. The author selected 100 of those subjects and obtained these results. Among the 53 females, the mean height is 63.7 in. and the standard deviation is 4.5 in.: among the 47 males, the mean height is 69.2 in. and the standard deviation is 3.1 in. Test the claim that the mean height of an adult California female is greater than the mean height of an adult California male when...
in a time use study, 20 randomly selected managers were found to spend a mean if...
in a time use study, 20 randomly selected managers were found to spend a mean if 2.25 hours each day on paperwork. the standard deviation of the 20 scores is s =1.7 hours. construct he 98% confidence Interval for the population standard deviation. use x2- distribution for analysis.
In a study of annual salaries of employees, random samples were selected from two companies to...
In a study of annual salaries of employees, random samples were selected from two companies to test if there is a difference in average salaries. For Company "X", the sample was size 65, the sample mean was $47,000 and the population standard deviation is assumed to be $11,000. For Company "Y", the sample size was 55, the sample mean was $44,000 and the population standard deviation is assumed to be $10,000. Test for a difference in average salaries at a...
1.) What is the probability that any randomly selected student (male or female) had a high...
1.) What is the probability that any randomly selected student (male or female) had a high school GPA between 3.00 and 4.00? Solve this problem using the Standard Normal Table (Z table). Show all work and provide the probability as a decimal rounded to four decimal places. 2.) If a female student had a high school GPA of 4.00, what percentile would this be for all female students? Solve this problem using the Standard Normal Table (Z table). Show all...
At a college, 61 female students were randomly selected and it was found that their monthly income had a standard deviation of $230.85. For 121 male students, the standard deviation was $320.53. Test the claim that variance of monthly incomes is higher for male students than it is foi female students. Use a = 0.01.
A store owner surveyed 25 randomly selected customers and found the ages shown (in years). The mean is 27.8 and the standard deviation is 8.72. The owner wants to know if the mean age of all customers is 32 years old. Use the given information to complete parts a through f 30 19 23 26 23 D 31 30 23 11 50 30 20 29 18 24 29 3740 46 35 26 23 22 27 23 a) What is the...
A store owner surveyed 25 randomly selected customers and found the ages shown (in years). The mean is 32.36 and the standard deviation is 11.79. The owner wants to know f the mean age of all customers is 36 years old. Use the given information to complete parts a through . 14 12 18 22 16 42 48 44 28 40 41 3743 44 38 16 20 39 48 31 30 34 21 47 36 a) What is the null...
The California Health Survey involved 51,048 randomly selected adults in California. The author selected 100 of those subjects and obtained these results. Among the 53 females, the mean height is 63.7 in. and the standard deviation is 4.5 in.: among the 47 males, the mean height is 69.2 in. and the standard deviation is 3.1 in. Test the claim that the mean height of an adult California female is greater than the mean height of an adult California male when...
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