Question

In: Statistics and Probability

A random sample of rain collected over 12 rain-days yielded a sample standard deviation of s...

A random sample of rain collected over 12 rain-days yielded a sample standard deviation of s = .319. A normal probability plot suggests the data comes from a population that is normally distributed; moreover, it appears no outliers are present. A 90% confidence interval for the sample standard deviation is sought. With the knowledge of the chi-square values derived from the question directly above, determine the confidence interval for the population standard deviation. Choose the best answer. Group of answer choices ( .266 , .542 ) ( .057 , .245 ) ( .239 , .495 ) ( .231 , .463 )

Solutions

Expert Solution

Solution :

Given that

2R = 2/2,df = 19.675

2L = 21 - /2,df = 4.575

The 90% confidence interval for is,

(n - 1)s2 / 2/2 < < (n - 1)s2 / 21 - /2

  11 * 0.319 2 / 19.675 < < 11 * 0.319 2 / 4.575

0.239 < < 0.495

(0.239 , 0.495)


Related Solutions

a random sample of 100 observations from a population. with standard deviation 60 yielded a sample...
a random sample of 100 observations from a population. with standard deviation 60 yielded a sample mean of 110. A. test the null hypothesis that m = 100 and the alternative hypothesis m > 100 using alpha = .05 and interpret the results b. test the null against the alternative hypothesis that m isn't equal to 110. using alpha = .05 and interpret the results c. compare the p values of the two tests you conducted. Explain why the results...
a random sample of 100 observations from a population. with standard deviation 60 yielded a sample...
a random sample of 100 observations from a population. with standard deviation 60 yielded a sample mean of 110. A. test the null hypothesis that m = 100 and the alternative hypothesis m > 100 using alpha = .05 and interpret the results b. test the null against the alternative hypothesis that m isn't equal to 110. using alpha = .05 and interpret the results c. compare the p values of the two tests you conducted. Explain why the results...
A random sample of 100 observations from a population with standard deviation 17.83 yielded a sample...
A random sample of 100 observations from a population with standard deviation 17.83 yielded a sample mean of 93. 1. Given that the null hypothesis is ?=90 and the alternative hypothesis is ?>90 using ?=.05, find the following: (a) Test statistic = (b) P - value: (c) The conclusion for this test is: A. There is insufficient evidence to reject the null hypothesis B. Reject the null hypothesis C. None of the above 2. Given that the null hypothesis is...
A random sample of 100 observations from a population with standard deviation 23.99 yielded a sample...
A random sample of 100 observations from a population with standard deviation 23.99 yielded a sample mean of 94.1 1. Given that the null hypothesis is μ=90μ=90 and the alternative hypothesis is μ>90μ>90 using α=.05α=.05, find the following: (a) Test statistic == (b) P - value: (c) The conclusion for this test is: A. There is insufficient evidence to reject the null hypothesis B. Reject the null hypothesis C. None of the above 2. Given that the null hypothesis is...
A random sample of 100 observations from a population with standard deviation 14.29 yielded a sample...
A random sample of 100 observations from a population with standard deviation 14.29 yielded a sample mean of 92.7. 1. Given that the null hypothesis is μ=90 and the alternative hypothesis is μ>90 using α=.05, find the following: (a) Test statistic = (b) P - value: (c) The conclusion for this test is: A. There is insufficient evidence to reject the null hypothesis B. Reject the null hypothesis C. None of the above 2. Given that the null hypothesis is...
A random sample of 100 observations from a population with standard deviation 70 yielded a sample...
A random sample of 100 observations from a population with standard deviation 70 yielded a sample mean of 113. Complete parts a through c below. a. Test the null hypothesis that muequals100 against the alternative hypothesis that mugreater than​100, using alphaequals0.05. Interpret the results of the test. What is the value of the test​ statistic? b. test the null hypothesis that mu = 100 against the alternative hypothesis that mu does not equal 100, using alpha=.05. interpret the results of...
A random sample of 100 observations from a population with standard deviation 22.99 yielded a sample...
A random sample of 100 observations from a population with standard deviation 22.99 yielded a sample mean of 94.1. 1. Given that the null hypothesis is μ≤90 and the alternative hypothesis is μ>90 using α=.05, find the following: (a) Test statistic = (b) P - value: (c) The decision for this test is: A. Fail to reject the null hypothesis B. Reject the null hypothesis C. None of the above 2. Given that the null hypothesis is μ=90 and the...
the following represents the PH of rain for a random sample of 12 rain dates in...
the following represents the PH of rain for a random sample of 12 rain dates in a particular region. A normal probability distribution plot suggest that the data could come from a population that is normally distributed. a boxplot indicates there are no outliers. The sample standard deviation s=0.326. construct and inerprete a 99% confidence interval for the standard deviation PH of rain water in this region
(1 point) A random sample of 100100 observations from a population with standard deviation 19.788150778587319.7881507785873 yielded...
(1 point) A random sample of 100100 observations from a population with standard deviation 19.788150778587319.7881507785873 yielded a sample mean of 93.893.8. (a)    Given that the null hypothesis is ?=90μ=90 and the alternative hypothesis is ?>90μ>90 using ?=.05α=.05, find the following: (i)    critical z/t score     equation editor Equation Editor (ii)    test statistic == (b)    Given that the null hypothesis is ?=90μ=90 and the alternative hypothesis is ?≠90μ≠90 using ?=.05α=.05, find the following: (i)    the positive critical z/t score     (ii)    the negative critical z/t score     (iii)    test statistic ==...
The following data represent the pH of rain for a random sample of 12 rain dates....
The following data represent the pH of rain for a random sample of 12 rain dates. A normal probability plot suggests the data could come from a population that is normally distributed. A boxplot indicates there are no outliers. Complete parts​ (a) through​ (d) below. 5.30 5.72 4.99 4.80 5.02 4.57 4.74 5.19 5.29 4.76 4.56 4.91 ​(a) Determine a point estimate for the population mean. A point estimate for the population mean is nothing. ​(Round to two decimal places...
ADVERTISEMENT
ADVERTISEMENT
ADVERTISEMENT