Consider the e-billing case. The mean and the standard deviation of the sample of n = 65 payment times are ē = 18.2779 and s = 3.9045. Test H9: µ = 19.0 versus Ha: µ < 19.0 by setting a equal to .01 and using a critical value rule and assume normality of the population. (Round your "t" and "to.01" answers to 3 decimal places and p-value answer to 4 decimal places. Negative values should be indicated by a minus sign. Use a statistical software package - e.g., Minitab, MegaStat, etc., to derive the p-value.) t0.01 p-value

Emissions of sulfur dioxide set off chemical changes in the atmosphere that result in "acid rain." The acidity of liquids is measured by pH on a scale of 0 to 14. Distilled water has pH 7.0, and lower pH values indicate acidity. Normal rain is somewhat acidic, so acid rain is sometimes defined as rainfall with a pH below 5.0.    Suppose the pH of rain at one location varies among rainy days according to a Normal distribution with mean 5.30 and standard deviation 0.43. What proportion of rainy days have rainfall with pH below 5.0?

From generation to generation, the mean age when smokers first start to smoke varies. However, the standard deviation of that age remains constant at around 2.1 years. A survey of 41 smokers of this generation was done to see if the mean starting age is at least 19. The sample mean was 18.2 with a sample standard deviation of 1.3. Do the data support the claim at the 5% level? Note: If you are using a Student's t-distribution for the problem, you may assume that the underlying population is normally distributed. (In general, you must first prove that assumption, though.) Part 1) Sketch a picture of this situation. Label and scale the horizontal axis and shade the region(s) corresponding to the p-value Part 2) Alpha (Enter an exact number as an integer, fraction, or decimal.)α =  Part 3) Construct a 95% confidence interval for the true mean. Sketch the graph of the situation. Label the point estimate and the lower and upper bounds of the confidence interval. (Round your lower and upper…

In a sample of 14 randomly selected high school seniors, the mean score on a standardized test was 1184 and the standard deviation was 162.8. Further research suggests that the population mean score on this test for high school seniors is 1014. Does the t-value for the original sample fall between - to 99 and to 99? Assume that the population of test scores for high school seniors is normally distributed. The t-value of t = fall between to.99 and to.99 because to.99 %3D (Round to two decimal places as needed.)