1) For the following data values below, construct a 90% confidence interval if the sample mean...

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Statistics

1) For the following data values below, construct a 90%confidence interval if the sample mean is known to be 0.719 and thestandard deviation is 0.366. (Round to the nearest thousandth)(Type your answer in using parentheses! Use a comma when inputingyour answers! Do not type any unnecessary spaces! List your answersin ascending order!) for example: (0.45,0.78) 0.56, 0.75, 0.10,0.95, 1.25, 0.54, 0.88

2) For the following data values below, construct a 98%confidence interval if the sample mean is known to be 9.808 and thepopulation standard deviation is 5.013. (Round to the nearestthousandth) (Type your answer in using parentheses! Use a commawhen inputing your answers! Do not type any unnecessary spaces!List your answers in ascending order!) for example: (0.45,0.78)6.6, 2.2, 18.5, 7.0, 13.7, 5.4, 5.3, 5.9, 4.7, 14.5 2.0, 14.8, 8.1,18.6, 4.5, 17.7, 15.9, 15.1, 8.6, 5.2 15.3, 5.6, 10.0, 8.2, 8.3,9.9, 13.7, 8.5, 8.2, 7.9 17.2, 6.1, 13.7, 5.7, 6.0, 17.3, 4.2,14.7, 15.2, 3.3 3.2, 9.1, 8.0, 18.9, 14.2, 5.1, 5.7, 16.4, 10.1,6.4

3)In a randomized controlled trial in Kenya, insecticide treatedbednets were tested as a way to reduce malaria. Among 343 infantsusing bednets, 15 developed malaria. Among 294 infants not usingbednets, 27 developed malaria. Want to use a 0.01 significancelevel to test the claim that the incidence of malaria is lower forinfants using bednets. Find the test statistic. (Round to thenearest hundredth) (ONLY TYPE IN THE NUMBER!)

4)State the conclusion whether or not to REJECT or FAIL TOREJECT the null hypotheses.(Check your spelling!) The originalclaim: The percentage of of M&Ms is greater than 5%. Thehypothesis test results in a p-value of 0.0010. a=0.05

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3.6 Ratings (291 Votes)
1Since we do not know the population standard deviation wewill use t statistic to construct 90 confidence intervalDegree of freedom n 1 7 1 6Critical value of t at 90 confidence interval and df 6 is194Margin of error t standard deviation 194 0366 07190 confidence    See Answer
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