The mean SAT score in mathematics is 550. The standard deviation of these scores is 30.A special preparation course claims that the mean SAT score, µ, of itsgraduates is greater than 550. An independent researcher tests this by taking a random sample of 14 students who completed the course; the mean SAT scorein mathematics for the sample was 555. Assume that the population is normally distributed. At the 0.05 level of significance, can we conclude that thepopulation mean SAT score for graduates of the course is greater than 550? Assume that the population standard deviation of the scores of course graduates isalso 30.Perform a one-tailed test. Then complete the parts below.Carry your intermediate computations to three or more decimal places, and round your responses as specified below. (If necessary, consult a list of formulas.)(a) State the null hypothesis H, and the alternative hypothesis H,.H, :0H :0(b) Determine the type of test statistic to use.(Choose one)ND=D0(c) Find the value of the test statistic. (Round to three or more decimal places.)O (a) State the null hypothesis H, and the alternative hypothesis H,.Ho:文H,:](b) Determine the type of test statistic to use.(Choose one) ▼コーロロSロロ20(c) Find the value of the test statistic. (Round to three or more decimal places.)D#0 OO(d) Find the critical value. (Round to three or more decimal places.)(e) Can we support the preparation course's claim that the population mean SATscore of its graduates is greater than 550?O Yes ONo

a) Assume that μ is the population mean SAT scores for graduates of the course.It is needed to check whether the population mean SAT scores for graduates of the course is greater than 550. The null and alternative hypotheses are, Null hypothesis: H0: μ = 550 Alternative hypothesis: H1: μ > 550 The direction of the test is right tailed. b) In this case, the population standard deviation is given and the population is normally distributed. Hence, one can use z test. c) The population standard deviation is σ = 30. The sample size is, n = 14. The sample mean is x̄ = 555. The z-test statistic is: z = (x̄ – μ)/ (σ/√n) = (555 – 550)/(30/√14) = 5/8.017837257 ≈ 0.6236 The test statistic is 0.6236. d) The level of significance is α = 0.05. The zα is the critical value of the standard normal distribution, above which, 100(α)% or α proportion of the observations lie. That is, α = 0.05 and 1 – α = 0.95. It implies that above z0.05, 5% of the observations lie and below z0.05, 95% (= 100 – 5) of the observations lie. As a result, from a standard normal table corresponding to row of 1.6 and column of .04 or using the excel formula =NORM.S.INV(0.95), it is obtained that, P(z < 1.645) = 0.95 or P(z > 1.645) = 0.05. Thus, the critical value is z0.05 = 1.645 and the rejection region is z > 1.645. e) The test statistic 0.6236 is less than the crucial value of 1.645. That is, the test statistic does not lie in the rejection region. At the α = 0.05 level, we fail to reject the null hypothesis and conclude the data are not statistically significant. Thus, no, we can not support the preparation course’s claim that the population mean SAT score of its graduates is greater than 550.