Find the P-value for the indicated hypothesis test with the given standardized test statistic, zDecide whether to reject H_{0} for the given level of significance Two-tailed test with test statistic z = - 1.5 and alpha = 0.08

Use dataset "HT_4143.xls" and conduct a two-sample t-test using Excel. You'd like to test the null hypothesis that the means of the two samples (column A and column B) are the same. The alternative hypothesis is that they are not the same. You have no reason to believe that the standard deviations of the two samples are equal.  Test at the alpha = 0.10 level.   After using Excel, what do you conclude? Are the means the same? Group of answer choices You reject the null hypothesis. Therefore, you conclude that the means of the two populations are different. You reject the null hypothesis. Therefore, you conclude that the means of the two populations are the same. You cannot reject the null hypothesis. Therefore, you conclude that the means of the two populations are different. You cannot reject the null hypothesis. Therefore, you conclude that the means of the two populations are the same.   HT_ 4143.xls: X1          X299.73     98.6995.93     98.9494.92     96.63100.52   99.16101.28…

Consider a drug testing company that provides a test for marijuana usage. Among 266 tested​ subjects, results from 30 subjects were wrong​ (either a false positive or a false​ negative). Use a 0.01 significance level to test the claim that less than 10percent of the test results are wrong.     Identify the null and alternative hypotheses for this test. Choose the correct answer below. test statistic__ pvalue___ identify the conclusion for this hypothesis test

in a one-tail hypothesis test where H_{0} is only rejected in the upper tail, the p-value 0.1029 and Z STAT =+1.27 what is the statistical decision if the null hypothesis is tested at the 0.08 level of significance? What is the statistical decision? H_{0} Since the p-value is (Type an integer or a decimal. Do not round.

1)TikTok users spend an average of 55 minutes per day using the app.  A researcher believes the time spend is less among teenager users. A random sample of 61 TikTok teenager users resulted in a sample mean daily use of 50 minutes, and a sample standard deviation of 12.5 minutes. Is there enough evidence to support the researcher’s claim? Use a level of significance of 5%. PLEASE SHOW ALL WORK FOR ALLL PARTS AND EXPLAIN THE ANSWER SO I CAN UNDERSTAND BETTER THANK YOU! a.)The null hypothesis is: . b.)The alternative hypothesis is:  c.)The test statistic is:  d.)The critical value is:  e.)Can you reject the null hypothesis? Yes or No.  f.)State your conclusion in the context of the problem:

A random sample of 777 subjects was asked to identify the day of the week that is best for quality family time. Consider the claim that the days of the week are selected with a uniform distribution so that all days have the same chance of being selected. The table below shows goodness-of-fit test results from the claim and data from the study. Test that claim using either the critical value method or the P-value method with an assumed significance level of a = 0.05. Test statistic, Num Categories 7 9.724 Degrees of freedom 6 Critical x? 12.592 Expected Freq 111.0000 P-Value 0.1368 ... Determine the null and alternative hypotheses. Họ: H,: Identify the test statistic. * = (Type an integer or a decimal.) %3D

A home improvement company is interested in improving customer satisfaction rate from the 71% currently claimed. The company sponsored a survey of 224 customers and found that 171 customers were satisfied. Determine whether sufficient evidence exists that the customer satisfaction rate is different than the claim by the company.     What is the test statistic z? What is the p-value? Does sufficient evidence exist that the customer satisfaction rate is different than the claim by the company at a significance level of �=0.1?

The drug Eliquis is used to help prevent blood clots in certain patients. In a clinical trial of 5924 patients treated withEliquis, 153 developed the adverse reaction ofnausea. Test the claim that 3% of Eliquis users develop nausea. Use a two tailed test of hypothesis,and the pvalue methodata 90% confidence level.Use a left tailed test of hypothesis, and the critical values method to test the claimat a 99% confidence level.

It has long been stated that the mean temperature of humans is 98.6°F. However, two researchers currently involved in the subject thought that the mean temperature of humans is less than 98 6°F They measured the temperatures of 44 healthy adults 1 to 4 times daily for 3 days, obtaining 200 measurements The sample data resulted in a sample mean of 982°F and a sample standard deviation of 0.9°F Use the P-value approach to conduct a hypothesis test to judge whether the mean temperature of humans is less than 98.6°F at the a=0.01 level of significance State the hypotheses. Ho 98.6 F 98 6 F

A simple random sample of front seat occupants involved in car crashes is obtained. Among 2780 occupants not wearing seat belts, 37 were killed. Among 7731 occupants wearing seat belts, 15 were killed. Use a 0.05 significance level to test the claim that seat belts are effective in reducing fatalities. (a) Test the claim using a hypothesis test. Ho: P1 = P2 H1: Choose one of the following letter_ A: P1 = P2: B: P1 > P2: C: P1 < P2 D: P1 + P2 (b) Identify the test statistic. z test = (Round to two decimal places.) %3D (c) Identify the P-value. Conclusion p-value is (Round to three decimal places.) since p-value is [ less than, or greater than) s.f.0.05 so ( Reject / Fail to reject. ) (d) Test the claim by constructing an appropriate confidence interval. < Pi – P2 << (Round to three decimal places.)