QUESTION 8A statistics student randomly selected 45 packets of sugar and weighed the contents of each packet, getting a mean of3.586 g and a standard deviation of 0.194 g. Test the claim that the weights of the sugar packets have a mean no longerequal to 3.5 g, as indicated on the label. Let "alpha" be 0.10 to test this claim.What is the decision/summary statement for this test?OA Reject the Null Hypothesis. There is enough evidence to support the claim that the sugar packets have a mean nolonger equal to 3.5 grams.O B. Reject the Null Hypothesis. There is enough evidence to reject the claim that the sugar packets have a mean nolonger equal to 3.5 grams.O. Do not reject the Null Hypothesis. There is not enough evidence to reject the claim that the sugar packets have amean no longer equal to 3.5 grams.OD. Do Not Reject the Null Hypothesis. There is not enough evidence to support the claim that the sugar packets have amean no longer equal to 3.5 grams.QUESTION 9Among 200 randomly selected Internet users, it was found that 89 of them use the Internet for making travel plans. Use asignificance level of alpha equal to 0.01 to test the claim that at least 50% of Internet users use it for making travel plans.What are the hypotheses?OA Ho: p 2 0.445H.n<0 445Click Save and Submit to save and submit. Click Save All Answers to save all answers.P Type here to search4-4+13ORF10PriscF12Insert234.8%3DERTTYUP

Note: Hi there! Thank you for posting the question. As there are several independent questions, according to our policy we have solved the first question. If you need help with other question, please re post separately.Denote μ as the population mean contents of each packet. It is needed to test the claim that the weights of the sugar packets have a mean no longer equal to 3.5 g. The null and alternative hypotheses are, Null hypothesis: H0: μ = 3.5. Alternative hypothesis: H1: μ ≠ 3.5 (claim) The direction of the test is both tailed. Thus, the test is the two-tailed test. In this case, the sample size is large enough (more than 30). The population standard deviation is not known. Hence, two-tailed t test should be performed. It is given that x̅ = 3.586 is the sample mean, μ is the population mean, s = 0.194 is the sample standard deviation and n = 45 is the sample size. The degrees of freedom (df) for a one-sample t-test, for a sample of size n is (n – 1). Here, n = 45, so that the number of degrees of freedom is df = 44 (= 45 – 1). The test statistic is: t = (x̅ – μ) / (s/√n). Thus, t = (3.586 – 3.5) / (0.194/√45) = 0.086/0.028919812 ≈ 2.97 The test statistic is 2.97. Now, using the EXCEL formula =T.DIST.2T(2.97,45) the p-value is obtained as approximately 0.0048. Consider the level of significance is α = 0.10. Decision rule: If p-value is less than or equal to the level of significance, then reject the null hypothesis. Otherwise, failed to reject the null hypothesis. Conclusion: The p-value is less than α = 0.10. Reject the Null Hypothesis. There is enough evidence to support the claim that the sugar packets have a mean no longer equal to 3.5 grams.