(ii) Test the claim at a 5% significance level using a critical value approach.
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- Spam: A researcher reported that 71.8 % of all email sent in a recent month was spam. A system manager at a large corporation believes that the percentage at his company may be 69%. He examines a random sample of 500 emails received at an email server, and finds that 365 of the messages are spam. Can you conclude that greater than 69% of emails are spam? Use both a=0.01 and a= 0.05 levels of significance and the critical value method with the table. Part 1 of 5 State the appropriate null and alternate hypotheses. Ho: P- .69 H1: p> .69 This hypothesis test is a right-tailed V test. Part 2 of 5 Find the critical values. Round the answers to three decimal places. For a=0.01 , the critical value is 2.326 For a=0.05 , the critical value is 1.645 Part: 2/5 Part 3 of 5 Compute the test statistic. Do not round intermediate calculations. Round the answer to two decimal places.A political science professor wants to determine her students’ attitude toward a presidential executive order preventing the construction of the Keystone Natural Gas pipeline. She has a sample of 50 students complete an interval level, 7-point scale assessing their attitude toward the policy where 1 = highly unfavorable 2= moderately unfavorable, 3 = slightly unfavorable, 4 = neutral, 5 = slightly favorable, 6 = moderately favorable, 7 = highly favorable. The sample mean is 4.6 with a standard deviation of 1.2. Using α = .01 what can she conclude regarding her students’ attitude toward the policy?Is this p-value significant at the 10% significance level? Is it significant at the 5% significance level? Compare the answers to these questions to your answers to parts (g) and (i) in two complete sentences.
- (b) Is there sufficient evidence to conclude, at the 5% level of significance, that the false positive rate is less than 5%? Explain.In a test of HO: p = 0.50 vs. HA: p > 0.50, which of the following describes the decision rule at the 5% level of significance. I. Reject HO if the p-value is more than 0.05. II. Reject HO if the p-value is less than 0.05. II. Reject HO if the test statistic is more than 0. %3DTo determine diet quality, male weanling rats were fed diets with various protein levels. Each of 15 rats was randomly assigned to one of three diets, and their weight gain in grams was recorded in the following table: Diet protein level Medium Low High 9.72 8.54 9.19 9.67 9.32 11.11 8.15 8.76 14.45 6.75 9.30 10.39 9.55 10.45 12.16 Sample Mean Sample Variance 1.696 8.768 9.274 11.460 0.547 3.963 Suppose we fit the data with the following model: Xij = µ+a; + Eij, i = 1, 2, 3; j= 1,., 5, N(0, o²) independently. Define X;. i = 1,2, 3, X. = 1=1 Xij, and SS(E) = E-1E=1(Xij – X;.)². E- Xij for where a1 + a2 + az = 0 and Eij Some R output that may help. > p1 qf (p1, 2, 12) [1] 0.010 0.025 0.052 0.106 2.807 3.885 5.096 6.927 > qf (p1, 2, 14) [1] 0.010 0.025 0.051 O.106 2.726 3.739 4.857 6.515 > qf (p1, 3, 12) [1] 0.037 0.070 0.114 O.192 2.606 3.490 4.474 5.953 > qf (p1, 3, 14) [1] 0.037 0.070 0.115 0.192 2.522 3.344 4.242 5.564
- A marketing survey involves product recognition in New York and California. Of 36 New Yorkers surveyed, 201 knew the product while 97 out of 306 Californians knew the product. At the 7% significance level, use the critical value method to test the claim that the recognition rates are the same in both states. Enter the smallest critical value. (Round your answer to nearest thousandth.)An immunologist is testing the hypothesis that the current flu vaccine is less than 71% effective against the flu virus. The immunologist is using a 10% significance level and thesehypotheses: H0: p=0.71 and Ha: p<0.71. Explain what the 10% significance level means in context.Q12
- Nutritional researchers conducted an investigation of two high-fiber diets intended to reduce cholesterol level. a group of people with high cholesterol were randomly selected to receive an "rice" diet or a "vegetable" diet for 20 days. Use a t-test to compare the diets at 5% significance level. Null Hypothesis: The means from two respective diets are the same Alternative Hypothesis: The means from the two diets differ. Fall in Cholesterol (mg/dL) Diet n Mean SD Rice 10 14.4 6.8 Vegetable 9 10.57 6.37 do we fail to reject the null hypothesis or reject null hypothesisFind the critical value in the following situations: a). T-test, left tail, level of significance 5%, degree of freedom 8 b). T-test, two tail, level of significance 20%, degree of freedom 22