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Secure l https://www mathalcom/Student/Player Testaspa?testld a1000363868 centerwin ST210-802 Statistical Reasoning and Applications Test: ST 210 Spring 2017 Final Exam Tine Remaining 0111 as StmtTest This Question: 3 pts 16 of 20 (5 complete This Test 60 pts possible E Question Hep A college entrance exam company determined that a score of 20 onthe mathematics portion of the exam mathematics. To achieve this goal, the company recommends that students take a core of math courses inihigh school Suppose amdam sample of 150 suggest who this se exam of 33 Do these results that students who complete the core cuniculum are ready for colegelevel mathematics? That is, are they sconng above20 on manportion of the exam? Complete parts a through d a State the appropriate nul and atemative hypotheses choose the corect answer below O A H. 20 versus H 20 OB p 204 versus p 204 O C Ho 204 versus H1 204 O D. H 20 versus H P 20 O E. H. w 204 versus H, 204 b) Verify that the requinements to perform the test using the -distribution are satisfied Isthe sample obtained using simple random samping or from a randomized experiment? A. Yes, because only high school students were sampled B. No, because only high school students were sampled o c Yes, because the students were randomly sampled O D. No, because not all students complete the courses is the population from which the sample is drawn normally distributed oris the sample size, n large (na 30 O A. Yes, the sample size is O B Yes, the population is nommaly distributed. O C No, neither of these conditions are true. Click to select your answer(s)Explanation / Answer
Solution:-
a) State the hypotheses. The first step is to state the null hypothesis and an alternative hypothesis.
Null hypothesis: = 20
Alternative hypothesis: > 20
Note that these hypotheses constitute a one-tailed test.
b) Yes becuase the students were randomly sampled.
Yes, the sample is larger than 30.
Yes, becuase the student's test score does not affect the other studnet's test score's.
c)
Formulate an analysis plan. For this analysis, the significance level is 0.10. The test method is a one-sample t-test.
Analyze sample data. Using sample data, we compute the standard error (SE), degrees of freedom (DF), and the t statistic test statistic (t).
SE = s / sqrt(n)
S.E = 0.2694
DF = n - 1 = 150 - 1
D.F = 149
t = (x - ) / SE
t = - 1.485
where s is the standard deviation of the sample, x is the sample mean, is the hypothesized population mean, and n is the sample size.
The observed sample mean produced a t statistic test statistic of 1.485. We use the t Distribution Calculator to find P(t > 1.485) = 0.0698
Interpret results. Since the P-value (0.0698) is less than the significance level (0.10), we have to reject the null hypothesis.
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