The test statistic for testing Họ: u = 100 against Ha: u ± 100 was t = 3.3, with P-value 0.001. Which of the following is the appropriate conclusion? Since the P-value is very small, we are unable to conclude that u = 100. In order to interpret the results, we must first know the sample size. Since the P-value is very small, we conclude that u# 100. Since the P-value is very small, we are unable to conclude that u # 100. Since the P-value is very small, we conclude that u = 100.

my answer for part b) of 6.24 was: (4.3, 9.26) so the variability in Hours is equal to 4.3 hours to 9.26 hours and in minutes its 258 to 555.6 minutes. I also posted 6.24 the previous exercise so that you have all the information but I need help with 6.25.

A new medication is being developed to treat severe nausea and vomiting among chemotherapy patients. The information below was generated from a small sample size (n = 11) and shows the number of times a chemotherapy patient vomited both before and after treatment due to the side effects of chemotherapy. Set up and interpret the results of a sign test at the alpha equals .05 level to determine if there is a difference in symptoms before and after treatment. Chemotherapy Patient Before Treatment After Treatment Difference Sign 1 9 3     2 12 5     3 14 7     4 9 4     5 8 2     6 11 3     7 15 7     8 12 8     9 7 8     10 9 5     11 8 7     A) We fail to reject H0, which states the median difference in the number of vomiting episodes in a day before and after treatment are equal because 1 is less or equal to critical value 1 for a sign test…

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Suppose IQ scores were obtained for 20 randomly selected sets of couples. The 20 pairs of measurements yield x = 101.02, y = 100.75, r = 0.852, P-value = 0.000, and y = 5.93 + 0.94x, where x represents the IQ score of the husband. Find the best predicted value of y given that the husband has an lQ of 100? Use a significance level of 0.05. %3D Click the icon to view the critical values of the Pearson correlation coefficient r. The best predicted value of y is Critical values of the pearson correlation coefficient r (Round to two decimal places as needed.) Critical Values of the Pearson Correlation Coefficient r X = 0.05 a = 0.01 INOTE: To test Ho: p=0 n Jagainst H,: p+0, reject Ho if the absolute value of r is greater than the critical value in the table. 4 0.950 0.990 0.878 0.959 0.811 0.917 7 0.754 0.875 0.707 0.834 0.666 0.798 10 0.632 0.765 11 0.602 0.735 12 0.576 0.708 13 0.553 0.684 14 0.532 0.661 15 0.514 0.641 16 0.497 0.623 17 0.482 0.606 18 0.468 0.590 19 0.456 0.575 20 0.444…

the data support the claim that the average ACT composite score of students from this students is randomly selected from a local high school with an X = 23 and s = 5.2. Do ACT composite scores are normally distributed with a u = 21. A sample of 25 students is randomly selected from a local high school with an X= 23 and s =5.L.D high school is different than 21 on the ACT composite score, using x = .05?

Use the data and table below to test the indicated claim about the means of two paired populations (matched pairs). Assume that the two samples are each simple random samples selected from normally distributed populations. Make sure you identify all values.     The table below shows the blood glucose of 20 students before breakfast and two hours after breakfast, using a specific insulin dosing formula to cover carbohydrates. Is there compelling statistical evidence that the specific insulin dosing formula is effective in reducing blood glucose levels? Use a significance level of 0.05. We have the differences (gain or loss), but we still need to compute the mean, standard deviation, and know the sample size for the differences (use Excel or Sheets for this computation). Once those are found, we can apply methods that are used for a 1 sample t-test.     Student Before Meal After Meal Difference 1 161 114 -47 2 196 88 -108 3 132 80 -52 4 175 96 -79 5 169 112 -57 6 138 137…