Homework Answers
Conclusion : There is not sufficient evidence to warrant the rejection of the claim that the library on average is 3.2 books per patron.
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A randomly selected sample of 6 persons who borrowed books at a library showed that they...
In a sample of 88 adults selected randomly from one town, it is found that 6...
In a sample of 88 adults selected randomly from one town, it is found that 6 of them have been exposed to a particular strain of the flu. At the 0.01 significance level, test the claim that the proportion of all adults in the town that have been exposed to this strain of the flue differs from the nationwide percentage of 8%. H0: p = 0.08 HA: p ≠ 0.08. α = 0.01 Test statistic: z = -0.41. P-Value =...
a poll of 2094 randomly selected adults showed that 94% of them own cell phones A...
a poll of 2094 randomly selected adults showed that 94% of
them own cell phones
A poll of 2,094 randomly selected adults showed that 94% of them own cell phones. The technology display below results from a test of the claim that 92% of adults own cell phones. Use the normal distribution as an approximation to the binomial distribution, and assume a 0.01 significance level to complete parts (a) through (e). Test of p = 0.92 vs p*0.92 Samplex N...
7. A study of people who refused to answer survey questions provided the randomly selected sample...
7. A study of people who refused to answer survey questions
provided the randomly selected sample data shown in the table
below. At a = .01, test the claim the cooperation of the subject
response or refusal is independent of age category.
Review for Last Test 6. A study from a university medical center examined whether the risk of hepatitis C as related to whether people had tattoos and to where they got their tattoos. Do the data indicate that...
Pinworm: In Sludge County, a sample of 50 randomly selected citizens were tested for pinworm. Of...
Pinworm: In Sludge County, a sample of 50 randomly selected citizens were tested for pinworm. Of these, 11 tested positive. The CDC reports that the U.S. average pinworm infection rate is 12%. Test the claim that Sludge County has a pinworm infection rate that is greater than the national average. Use a 0.01 significance level. (a) What is the sample proportion of Sludge County residents with pinworm? Round your answer to 3 decimal places. p̂ = (b) What is the...
A poll of 2,084 randomly selected adults showed that 94% of them own cell phones. The...
A poll of 2,084 randomly selected adults showed that 94% of them own cell phones. The technology display below ret from a test of the claim that 92% of adults own cell phones. Use the normal distribution as an approximation to the bin distribution, and assume a 0.01 significance level to complete parts (a) through (e). Test of p = 0.92 vs p+0.92 Z-Value P-value Sample p 95% CI N Sample X 0.000 4.01 (0.930869,0.956847) 1 1967 2,084 0.943858 a....
_ 9) In a sample of 10 randomly selected women, it was found that their mean height was 634 inches. From previous...
_ 9) In a sample of 10 randomly selected women, it was found that their mean height was 634 inches. From previous studies, it is assumed that the standard deviation, , 124 inches and that the population of height measurements is normally distributed a) Construct the confidence interval for the population mean height of women b) If the sample size was doubled to 20 women, what will be the effect on the confidence interval? 10) 10) The numbers of advertisements...
6. Confidence intervals A survey of 90 randomly selected families showed that 40 owned at least...
6. Confidence intervals A survey of 90 randomly selected families showed that 40 owned at least one television set, Find a 95% confidence interval for the proportion of all families that own at least one television set. a. b. A random sample of 20 automobiles has a pollution by-product release standard deviation of 2.3 ounces when 1 gallon of gasoline is used. Find a 90% confidence interval for the population standard deviation of the pollution by-product release. An irate patient...
(1 point) A sample of 6 measurments, randomly selected from a normally distributed population, resulted in...
(1 point) A sample of 6 measurments, randomly selected from a normally distributed population, resulted in a sample mean x = 7.5 and sample standard deviation s = 1.08. Using a = 0.01, test the null hypothesis that the mean of the population is 8.1 against the alternative hypothesis that the mean of the population is less than 8.1 by giving the following: (a) the degree of freedom (b) the critical t value (c) the test statistic The final conclusion...
2. The average number of cups of coffee drunk each day by a randomly selected sample...
2. The average number of cups of coffee drunk each day by a randomly selected sample of 25 students was 4.4. Assume the population standard deviation is 0.5 cups of coffee. Does the sample provide enough statistical evidence to support a claim that students drink 4 cups of coffee per day on average? Test the hypothesis at a 1% level of significance. Perform a 1-sample Z-test of mean to assess the sample evidence. A. State the null and alternative hypotheses...
(6 points) Ten randomly selected people took an IQ test A, and next day they took...
(6 points) Ten randomly selected people took an IQ test A, and next day they took a very similar IQ test B. Their scores are shown in the table below. Person ABCDEFGHIJ Test A 109|83|74 97 11693|92|124101105 Test B 114837497 1199690124 102 108 1. Consider (Test A - Test B). Use a 0.01 significance level to test the claim that people do better on the second test than they do on the first. (Note: You may wish to use software.)...
a poll of 2094 randomly selected adults showed that 94% of
them own cell phones
A poll of 2,094 randomly selected adults showed that 94% of them own cell phones. The technology display below results from a test of the claim that 92% of adults own cell phones. Use the normal distribution as an approximation to the binomial distribution, and assume a 0.01 significance level to complete parts (a) through (e). Test of p = 0.92 vs p*0.92 Samplex N...
7. A study of people who refused to answer survey questions
provided the randomly selected sample data shown in the table
below. At a = .01, test the claim the cooperation of the subject
response or refusal is independent of age category.
Review for Last Test 6. A study from a university medical center examined whether the risk of hepatitis C as related to whether people had tattoos and to where they got their tattoos. Do the data indicate that...
A poll of 2,084 randomly selected adults showed that 94% of them own cell phones. The technology display below ret from a test of the claim that 92% of adults own cell phones. Use the normal distribution as an approximation to the bin distribution, and assume a 0.01 significance level to complete parts (a) through (e). Test of p = 0.92 vs p+0.92 Z-Value P-value Sample p 95% CI N Sample X 0.000 4.01 (0.930869,0.956847) 1 1967 2,084 0.943858 a....
_ 9) In a sample of 10 randomly selected women, it was found that their mean height was 634 inches. From previous studies, it is assumed that the standard deviation, , 124 inches and that the population of height measurements is normally distributed a) Construct the confidence interval for the population mean height of women b) If the sample size was doubled to 20 women, what will be the effect on the confidence interval? 10) 10) The numbers of advertisements...
6. Confidence intervals A survey of 90 randomly selected families showed that 40 owned at least one television set, Find a 95% confidence interval for the proportion of all families that own at least one television set. a. b. A random sample of 20 automobiles has a pollution by-product release standard deviation of 2.3 ounces when 1 gallon of gasoline is used. Find a 90% confidence interval for the population standard deviation of the pollution by-product release. An irate patient...
(1 point) A sample of 6 measurments, randomly selected from a normally distributed population, resulted in a sample mean x = 7.5 and sample standard deviation s = 1.08. Using a = 0.01, test the null hypothesis that the mean of the population is 8.1 against the alternative hypothesis that the mean of the population is less than 8.1 by giving the following: (a) the degree of freedom (b) the critical t value (c) the test statistic The final conclusion...
(6 points) Ten randomly selected people took an IQ test A, and next day they took a very similar IQ test B. Their scores are shown in the table below. Person ABCDEFGHIJ Test A 109|83|74 97 11693|92|124101105 Test B 114837497 1199690124 102 108 1. Consider (Test A - Test B). Use a 0.01 significance level to test the claim that people do better on the second test than they do on the first. (Note: You may wish to use software.)...
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