
1o 4 A random sample of 9 tourists in Chattanooga showed that they spent an of...
6. The average price of homes sold in the U.S. in 2012 was $240,000. A sample of 144 homes sold in Chattanooga in 2012 showed an average price of $246,000. It is known that the standard deviation of the population is $36,000. We are interested in determining whether or not the average price of homes sold in Chattanooga is significantly different than the national average. Let α = .01 a. Verify whether the Central Limit Theorem conditions are met. b....
The null and alternate hypotheses are: H0:µ1≤µ2. H0:µ1>µ2. A random sample of 29 items from the first population showed a mean of 112 and a standard deviation of 9. A sample of 15 items for the second population showed a mean of 97 and a standard deviation of 12. Use the .01 significance level. a. Find the degrees of freedom for unequal variance test b. State the decision rule for .1 significance level c. Compute the value of the test...
1. A random sample of 82 customers, who visited a department store, spent an average of $71 at this store. Suppose the standard deviation of expenditures at this store is O = $19. What is the e 98% confidence interval for the population mean? 2. A sample of 25 elements produced a mean of 123.4 and a standard deviation of 18.32 Assuming that the population has a normal distribution, what is the 90% confidence interval for the population mean? 3....
The null and alternate hypotheses are: H0: ?1 ? ?2 H1: ?1 > ?2 A random sample of 24 items from the first population showed a mean of 113 and a standard deviation of 13. A sample of 18 items for the second population showed a mean of 103 and a standard deviation of 14. A) Use the 0.05 significant level. Find the degrees of freedom for unequal variance test. (Round down your answer to the nearest whole number.) B)...
A random sample of 49 measurements from one population had a sample mean of 13, with sample standard deviation 3. An independent random sample of 64 measurements from a second population had a sample mean of 15, with sample standard deviation 4. Test the claim that the population means are different. Use the level of significance 0.01. Using s1 = 3 and s2 = 4, we can compute the t value corresponding to the test statistic x1 − x2 = −2. Recall...
A sample of 36 account balances of a credit company showed an average balance of $1,179 and a standard deviation of $136. You want to determine if the mean of all account balances is significantly greater than $1,150. Use a 0.05 level of significance. Assume the population of account balances is normally distributed. Compute the test statistic. A sample of 30 account balances of a credit company showed an average balance of $1,165 and a standard deviation of $125. You...
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Exercise 11-14 (LO11-2) The null and alternate hypotheses are: random sample of 27 tems from the first population showed a mean of 110 and a standard deviation of 15. A sample of 19 items for the second population showed a mean of 100 and a standard deviation of 6. Use the 0.025 significant level a. Find the de grees of freedom for unequal variance test. (Round down your answer to the nearest whole number) of freedom 21...
Each person in a random sample of 224 male teenagers and a random sample of 303 female teenagers was asked how many hours he or she spent online in a typical week. The sample mean and standard deviation were 15.2 hours and 11.3 hours for males and 14.3 and 11.7 for females. (Use a statistical computer package to calculate the P-value. Use μmales − μfemales. Round your test statistic to two decimal places, your df down to the nearest whole...
Each person in a random sample of 224 male teenagers and a random sample of 303 female teenagers was asked how many hours he or she spent online in a typical week. The sample mean and standard deviation were 15.2 hours and 11.3 hours for males and 14.3 and 11.7 for females. (Use a statistical computer package to calculate the P-value. Use μmales − μfemales. Round your test statistic to two decimal places, your df down to the nearest whole...
Question 10 (0.75 points) A study showed that in a certain month, the mean time spent per visit to Facebook was 19.5 minutes. Assume the standard deviation of the population is 8 minutes. Suppose that a simple random sample of 100 visits in that month has a sample mean of 21.13 minutes. A social scientist is interested in knowing whether the mean time of Facebook visits has increased. Perform the hypothesis test and compute the P-value. Write down your P-value....