The class that graduated from a college four years ago took an average of 2.3 science courses. Since then anew curriculum requirement has been introduced. The registrar is interested in the effect this requirementhas had on the enrollment in science courses. A random sample of the transcripts of 25 seniors found that themean number of science courses taken was 2.8 with a standard deviation of 1.01 . Do these results suggestthat the mean number of science courses taken has changed significantly since the new requirement wasintroduced? Test at a = 0.05.(a). The null and alternative hypotheses areO Ho : ã = 2.3Ha:ữ + 2.3Ο 1: μ = 2.8Ha: µ < 2.8Ho: H2.3O Ho:Ha:ã + 2.82.8Ha: µ + 2.3Part 2 of 5(b). The test statistic is(use 4 decimal places)

Here given sample statistics And significance level 0.05

Answered: (a). The null and alternative…

MATLAB: An Introduction with Applications

6th Edition

ISBN:9781119256830

Author:Amos Gilat

Publisher:Amos Gilat

Chapter1: Starting With Matlab

Section: Chapter Questions

Problem 1P

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An article includes the accompanying data on compression strength (lb) for a sample of 12-oz aluminum cans filled with strawberry drink and another sample filled with cola. Beverage Sample Size Sample Mean Sample SD Strawberry Drink 10 535 24 Cola 10 559 15 Does the data suggest that the extra carbonation of cola results in a higher average compression strength? Base your answer on a P-value. (Use ? = 0.05.) State the relevant hypotheses. (Use ?1 for the strawberry drink and ?2 for the cola.) H0: ?1 − ?2 = 0Ha: ?1 − ?2 ≥ 0H0: ?1 − ?2 = 0Ha: ?1 − ?2 ≠ 0 H0: ?1 − ?2 = 0Ha: ?1 − ?2 < 0H0: ?1 − ?2 = 0Ha: ?1 − ?2 > 0 Calculate the test statistic and determine the P-value. (Round your test statistic to one decimal place and your P-value to three decimal places.) t = P-value = State the conclusion in the problem context. Reject H0. The data suggests that cola has a higher average compression strength than the strawberry drink.Reject H0. The…
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(a). The null and alternative hypotheses are O Ho:ã = 2.3 Ha: + 2.3 Ο H0: μ = 2.8 На: д < 2.8 Ο Ηρ: μ = 2.3 Ha:µ # 2.3 O Ho:ã = 2.8 Ha:ã + 2.8 Part 2 of 5 (b). The test statistic is (use 4 decimal places)