The test statistic in a two-tailed test is z=0.67. Determine the P-value and decide whether, at the 10% significance level, the data provide sufficient evidence to reject the null hypothesis in favor of the alternative hypothesis. Click here to view a partial table of areas under the standard normal curve.
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- Which is not one of the major assumptions typically associated with parametric tests of significance? A. Random Selection B. Homogeneity of Variance C. Absence of restricted ranges1.List all of the characteristics of the Standard Normal Distribution. 2.Why is finding the z-score called standardizing data? 3.Does the Empirical Rule apply to the standard normal curve?Big babies: The National Health Statistics Reports described a study in which a sample of 310 one-year-old baby boys were weighed. Their mean weight was 24.4 pounds with standard deviation 5.3 pounds. A pediatrician claims that the mean weight of one-year-old boys is greater than 25 pounds. Do the data provide convincing evidence that the pediatrician's claim is true? Use the α = 0.10 level of significance and the P-value method with the TI-84 Plus calculator. Compute the P-value. Round the answer to at least four decimal places. P-value = ____ How do I solve this problem using the TI-84 calculator?
- purses Researchers conducted an experiment to test the effects of alcohol. Erors were recorded in a test of visual and motor skills for a treatment group of 21 people who drank ethanol and another group of 21 people given a placebo. The errors for the treatment group have a standard deviation of 2.10, and the errors for the placebo group have a standard deviation of 0.71. Use a 0.05 significance level to test the claim that the treatment group has errors that vary significantly more than the errors of the placebo group. Assume that the two populations are normally distributed. se Hom bus What are the null and alternative hypotheses? ndar O A. Ho: o =0 O B. Ho: of +o? ntation OC. Ho o =0 OD. Hoi o? =o} book puncem Identify the test statistic. (Round to two decimal places as needed.) gnments Use technology to identify the P-value. dy Plan (Round to three decimal places as needed.) debook What is the conclusion for this hypothesis test? O A. Reject Ho. There is sufficient evidence to…Americans are about 1 inch taller but nearly 25 pounds heavier than in 1960. The average BMI increased from approximately 25 to 28 in 2002. Boston is considered one of americas healthiest cities. Is the weight gain since 1960 similar in boston? Sample of n=25 and mean increase of 17 with standard deviation of 8.6. Is boston statistically significantly different in terms of weight gain since 1960? Run the appropriate test at 5%level of significance. A.state the null and the research hypotheses, b. Idebtify critical values of z 9r t used to define the decision rule, c. Calculate the test statistic, d. Report the p-value, e. Report your findingsThe test statistic of z = 1.98 is obtained when testing the claim that p ≠ 0.752. a. Identify the hypothesis test as being two-tailed, left-tailed, or right-tailed. b. Find the P-value. c. Using a significance level of a = 0.10, should we reject Ho or should we fail to reject Ho?
- The commercial for the new Meat Man Barbecue claims that it takes 9 minutes for assembly. A consumer advocate thinks that the assembly time is higher than 9 minutes. The advocate surveyed 14 randomly selected people who purchased the Meat Man Barbecue and found that their average time was 9.7 minutes. The standard deviation for this survey group was 2.2 minutes. What can be concluded at the the a = 0.05 level of significance level of significance? a. For this study, we should use z-test for a population proportion v b. The null and alternative hypotheses would be: Ho: H1: c. The test statistic (t 1.191 (please show your answer to 3 decimal places.) d. The p-value = 0.1169 (Please show your answer to 4 decimal places.) e. The p-value is ? ▼ a f. Based on this, we should | Select an answer the null hypothesis. g. Thus, the final conclusion is that ... The data suggest that the population mean amount of time to assemble the Meat Man barbecue is not significantly higher than 9 at a = 0.05,…Small-business telephone users were surveyed 6 months after access to carriers other than Carrier A became available for wide-area telephone service. Of a random sample of 240 Carrier A users, 135 said they were attempting to learn more about their options, as did 227 of an independent random sample of 260 users of alternate carriers. Test, at the 1% significance level against a two-sided alternative, the null hypothesis that the two population proportions are the same. Click the link to view a standard normal distribution table. be the Let Px be the proportion of Carrier A users who said they were attempting to learn more about their options and Py proportion of users of alternate carriers who said they were attempting to learn more about their options. Determine the null and alternative hypotheses. Choose the correct answer below. O A. Ho: Px-Py=0 H₁: Px-Py 20 O D. Ho: Px-P₁ #0 H₁: Px - Py = 0 B. Ho: Px-Py = 0 H₁: Px-Py 0Watching TV: In 2012, the General Social Survey asked a sample of 1318 people how much time they spent watching TV each day. The mean number of hours was 2.95 with a standard deviation of 2.63. A sociologist claims that people watch a mean of 3 hours of TV per day. Do the data provide sufficient evidence to conclude that the mean hours of TV watched per day differs from the claim? Use the a= 0.01 level of significance and the critical value method. Part: 0 / 5 Part 1 of 5 State the appropriate null and alternate hypotheses. H: H₁ : This hypothesis test is a (Choose one) test. 0<0 ☐☐ X □<口 μ 0=0 Ś
- The test statistic in a two-tailed test is z= 2.27. Determine the P-value and decide whether, at the 5% significance level, the data provide sufficient evidence to reject the null hypothesis in favor of the alternative hypothesis. E Click here to view a partial table of areas under the standard normal curve. The P-value is (Round to four decimal places as needed.)The average salary for American college graduates is $45,100. You suspect that the average is less for graduates from your college. The 50 randomly selected graduates from your college had an average salary of $42,788 and a standard deviation of $5,940. What can be concluded at the αα = 0.05 level of significance? For this study, we should use Select an answer t-test for a population mean z-test for a population proportion The null and alternative hypotheses would be: H0:H0: ? μ p Select an answer > < = ≠ H1:H1: ? p μ Select an answer ≠ > = < The test statistic ? z t = (please show your answer to 3 decimal places.) The p-value = (Please show your answer to 4 decimal places.) The p-value is ? > ≤ αα Based on this, we should Select an answer fail to reject reject accept the null hypothesis. Thus, the final conclusion is that ... The data suggest that the population mean is not significantly less than 45,100 at αα = 0.05, so there is…An article in the San Jose Mercury News stated that students in the California state university system take 6 years, on average, to finish their undergraduate degrees. A freshman student believes that the mean time is less and conducts a survey of 53 students. The student obtains a sample mean of 4.1 with a sample standard deviation of 1.9. Is there sufficient evidence to support the student's claim at an �=0.1 significance level? Test the claim: Determine the null and alternative hypotheses. Enter correct symbol and value.�0: �=Incorrect��: � Correct< Incorrect Determine the test statistic. Round to four decimal places.�=Incorrect Find the �-value. Round to 4 decimals.