The following MINITAB output presents the results of a hypothesis test for a population proportion p. Some of the numbers are missing. Fill in the numbers for (a) through (c). Test and CI for One Proportion: X Test of p = 0.7 vsp < 0.7 X N Sample p 95% Upper Bound z-Value P-Value 0.724021 Variable 345 500 (a) (b) (b) (c) х

a) which for these data is 32.1470, 42.3520, or 10.0674 c) that minimizes the error sum of squares, total sum of squares, or regression sum of squares…for which these data is 32.1470, 42.3520, or 10.0674

Our environment is very sensitive to the amount of ozone in the upper atmosphere. The level of ozone normally found is 4.4 parts/million (ppm). A researcher believes that the current ozone level is not at the normal level. The mean of 1090 samples is 4.3 ppm. Assume the variance is known to be 1.44. Does the data support the claim at the 0.05 level? Step 1 of 5: Enter the hypotheses: Step 2 of 5: Enter the value of the z test statistic. Round your answer to two decimal places. Step 3 of 5: Specify if the test is one-tailed or two-tailed. Step 4 of 5: Enter the decision rule. Step 5 of 5: Enter the conclusion.

a. For this study, we should use t-test for a population mean b. The null and alternative hypotheses would be: Ho: uv (Please enter a decimal) c. The test statistic z v (please show your answer to 3 decimal places.) %3D d. The p-value = (Please show your answer to 4 decimal places.) %3D e. The p-value is> va f. Based on this, we should fail to reject vthe null hypothesis. g. Thus, the final conclusion is that ...