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In 2018, the national average SAT score was 1067. Using the sample of 72 observations, we want to test whether the average SA
Fto-date With securit fx A C SAT Scores 1200 1194.58333 1200 226.049586 990 1350 1170 1020 900 1320 1100 1250 1110 1250 1100 the average of numbers and then Standard deviation
In 2018, the national average SAT score was 1067. Using the sample of 72 observations, we want to test whether the average SAT for ECO 2200 students is different from the national average. Given a .05 significance level, what is the test statistic? (Hint, in Excel, the standard error -STDEV.S/SQRT(n)) Select one: a. t 1.645 b. t 1.96 c. t 2.57 d. t 4.79 3 What do you conclude from the hypothesis test described in Question 22 Select one: ut of 1 a. Fail to Reject the Null b. Reject the Null c. Nullify the Rejection Region d. Accept the Aleternative W 44 F3 F2 & % 6 5 4 3 2 T R F W 7 OLO 10 +At lillL co AV
Fto-date With securit fx A C SAT Scores 1200 1194.58333 1200 226.049586 990 1350 1170 1020 900 1320 1100 1250 1110 1250 1100 1130 1170 7 1200 18 940 19 1030 20 1050 21 1070 22 23
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Answer #1

the .05 significance level in z table gives value of 1.645. If the value calculated by z test is greater than 1.645 we will reject the null hypothesis.Jn n May.8333 Tニ226-049S86 nニコ2- Pasteny aolues Z.9

Now the z value is 4.79 which is greater than 1.645 so we will reject the null hypothesis which is option B

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