Ten people went on a diet for a month. The weight losses experienced (in pounds) are given below. The negative weight loss is a weight gain. Test the hypothesis that the mea weight loss was more than 0, using a significance level of 0.05. Assume the population distribution is Normal. 3, 7, 10, 0, 4, 7, 5, 5, 1, and - 2 0 Determine the null and alternative hypotheses. Choose the correct answer below. O B. Ho: H=0 OC. Ho: H#0 Hạiµ= 0 O A. Ho: H<0 Ha: 20 Ha: H<0 Ο D . H : μ 0 O E. Ho: H>0 OF. Ho: H=0 Ha: u>0 Hg: us0 The test statistic is. (Round to two decimal places as needed.) The p-value is|. (Round to three decimal places as needed.) Interpret the results of the test. Since the p-value is less than or equal to the significance level, reject Ho. There is sufficient evidence to conclude that the population mean is more than 0 pounds at a significance level of 0.05.

The US Department of Energy reported that 48% of homes were heated by natural gas. A random sample of 331 homes in Oregon found that 132 were heated by natural gas. Test the claim that proportion of homes in Oregon that were heated by natural gas is different than what was reported. Use a 1% significance level. Give answer to at least 4 decimal places. What are the correct hypotheses? (Select the correct symbols and use decimal values not percentages.) H0:                   H1:                     Based on the hypotheses, compute the following:Test Statistic = p-value =  Based on the above we choose to       The correct summary would be:       that the proportion of homes in Oregon that were heated by natural gas is different than what the DOE reported value of 48%.

Use the information below, gathered from 1000 students from a small college who disclosed their age as well as their preference for type of instruction, to determine whether or not there is an association between age and preferred type of instruction. Write your hypotheses, perform the test, find the test statistic, the p-value, and state your conclusion. Assume a significance level of � = 0.01.       AGE       18 & under 19-25 26-35 36-45 65 or Older online 43 56 26 20 20 Scheduled Zoom 48 75 34 26 20 Hybrid Zoom 82 19 88 83 59 Face to Face 27 50 52 71 101 a) Write your hypotheses.b) Use your calculator to determine the test statistic and p-value.c) Using the significance level provided, write an appropriate conclusion

A teacher would like to determine if quiz scores improve after completion of a worksheet. The students take a pre-quiz before the worksheet and then another quiz after the worksheet. (Pre-quiz score - Post-quiz score) Assume quiz scores are normally distributed. The grades for each quiz are given below. Use a significance level of a = 0.05. H₂: Hd = 0 Ha: Hd <0 pre-quiz 12 16 17 16 11 15 12 7 15 13 post-quiz 13 16 15 16 14 15 13 9 14 15 What is the test statistic? test statistic = (Report answer accurate to 2 decimal places.) What is the p-value for this sample? p-value= (Report answer accurate to 3 decimal places.) The correct decision is to Select an answer Conclusion: Select an answer that the worksheet improved quiz scores.

Choose the appropriate statistical test. When computing, be sure to round each answer as indicated. A dentist wonders if depression affects ratings of tooth pain. In the general population, using a scale of 1-10 with higher values indicating more pain, the average pain rating for patients with toothaches is 6.8. A sample of 30 patients that show high levels of depression have an average pain rating of 7.1 (variance 0.8). What should the dentist determine? 1. Calculate the estimated standard error. (round to 3 decimals). [st.error] 2. What is thet-obtained? (round to 3 decimals). 3. What is the t-cv? (exact value) 4. What is your conclusion? Only type "Reject" or Retain"

Suppose IQ scores were obtained for 20 randomly selected sets of couples. The 20 pairs of measurements yield x= 100.1, y = 98.3, r=0.961, P-value = 0.000, and y = - 18.02 + 1.16x, where x represents the IQ score of the wife. Find the best predicted value of y given that the wife has an IQ of 100? Use a significance level of 0.05. Click the icon to view the critical values of the Pearson correlation coefficient r. The best predicted value of y is (Round to two decimal places as needed.)

Suppose IQ scores were obtained for 20 randomly selected sets of siblings. The 20 pairs of measurements yield x = 102.13, y = 101.35, r= 0.927, P-value = 0.000, and y = 2.05 + 0.97x, where x represents the IQ score of the older child. Find the best predicted value of y given that the older child has an IQ of 103? Use a significance level of 0.05. Click the icon to view the critical values of the Pearson correlation coefficient r. Critical Values of the Pearson Correlation Coefficient r The best predicted value of y is (Round to two decimal places as needed.) Critical Values of the Pearson Correlation Coefficient r a = 0.05 a = 0.01 NOTE: To test Ho: p=0 Jagainst H,: p#0, reject H, lif the absolute value of r is greater than the critical value in the table. 4. 0.950 0.990 0.878 0.959 6 0.811 0.917 0.754 0.875 8. 0.707 0.834 0.666 0.632 0.602 9 0.798 10 0.765 11 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 0.468 0.456 18 0.590 19 0.575…

The US Department of Energy reported that 48% of homes were heated by natural gas. A random sample of 317 homes in Oregon found that 166 were heated by natural gas. Test the claim that proportion of homes in Oregon that were heated by natural gas is different than what was reported. Use a 5% significance level. Give answer to at least 4 decimal places. What are the correct hypotheses? (Select the correct symbols and use decimal values not percentages.) Ho: P Test Statistic = 0° p-value = H₁: p Based on the hypotheses, compute the following: #v oº 48 48 <-Need answer, please. <-Need answer, please OB 0° Based on the above we choose to Fail to reject the null hypothesis ✓ OB The correct summary would be: There is not enough evidence to support the claim that the proportion of homes in Oregon that were heated by natural gas is different than what the DOE reported value of 48%.

A statistician changes her level of significance from .001 to .05.  What impact will this change have on her risk of making a Type I and Type II error?  Is the change in the level of significance a good decision?  Why or why not?