Final/QNT 561

| September 29, 2018

YOUR NAME Final/QNT 5611. A negative 0.947 correlation coefficient has been found between two variables after examining 43 pairs of observations. What can be concluded?a. There is a strong positive relationship between the two variables.b. There is a strong negative relationship between the two variables.c. There is a weak positive relationship between the two variables.d. There is a weak negative relationship between the two variables.2. A linear regression between Y and X produced the following equation for the least squares line:.png”>= –4.13 + 2.1xWhich of the following statements concerning this relationship is true?a. For every one-unit increase in X, Y increases 4.13 units.b. For every one-unit increase in X, Y decreases 2.1 units.c. For every one-unit increase in X, Y decreases 4.13 units.d. For every one-unit increase in X, Y increases 2.1 units.3- The following table shows the number of workdays absent based on the length of employment in years.Number of Workdays Absent 2 3 3 5 7 7 8Length of Employment (in yrs) 5 6 9 4 2 2 0Compute the Coefficient of Correlation, the R-Square, and the linear regression coefficients4. The price-to-book value is a commonly used measure of whether a stock is over- or underpriced. The average price-to-book value for all gas utility stocks averaged 1.80 in 2010 and the standard deviation was 0.45. A random selection of 36 gas utility stocks in August, 2011 yielded a mean of 1.95.a. Set up the null and alternative hypotheses to test if the average price-to-book value of the August, 1999 selection is greater than the 1998 national average for gas utility stocks.b. Determine the critical value usinga = 0.05c. Calculate the value of the test statisticzd. What is your conclusion?5. The e-mail usage for two different plants of a large company was compared at level of significance 0.05. Samples of 40 employees were selected at each plant. The mean number of e-mail messages sent per employee for one plant was 15.5 per week and the standard deviation was 5.0. For the other plant, the mean was 18.4 and the standard deviation was 1.6. For the test of equal population means versus unequal population meansa) state the null and alternate hypothesisb) Calculate the test statisticsc) Decide whether the population means are different.6- The following table shows the average number of daily e-mail messages that ACME Company received over a six-week measurement period.Week DayMTWTHFMessages5753524538George Warren, general manager, has always claimed that the day of the week has no bearing on the number of messages received.Using Chi-Square test of hypothesis methdolology:a) State the null and alternate hypothesisb) Calculate the value of the test statisticsc) At 5% significance level what is your decision?7- The following is a partial ANOVA table:Source SS df MS FTreatment ? 2 ? 8.00Error ? ? 20Total 500 11Calculate the missing numbers (?) in the table8-.png”>In 1999′ a sample of 200 in store shopper showed that 42 paid by debit card(a) Hypotheses:(b) Test:(c) p-value:The p-value corresponding to z = -2.2798 is

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