Question

A new ACT preparation they recorded the are normally distributed.) claims that high school and practice ACT scores of 6 individuals after a one-month period. The results are shown in the following table: (Assume that ACT scores Test to see whether the preparation program will increase the ACT sco Sa is 1.2 in the tablel) e on avernagye by more than 0as points (Note thut the samolrnot the sle ) Alternatives Select one: O Mo 0.85 H, a 0.85 O None of the other four (i) Test statistic Select one: 〇! 2.015 2.571 3.365 5.883
-3365 O None of the other four i) p-value Select one: O Approximately 0.01 O Approximately 0.02 O Approximately 0.025 O Approximately 0.05 O Approximately 0.10 (iv) What is the conclusion if the level of significance is a Select one: the ACT score by more than 0.85 points O Conclude Ho that the preparation program will NOT increase O Conclude Ho that the preparation program will ncrease the ACT score by more than o.85 points O Conclude H, that the preparation program will Increase the ACT score by more than 0.85 points O Conclude H, that the preparation program will NOT increase the ACT score by more than o.as points O None of the other four A Local
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Answer #1

(i)

Since this is a One-Tailed Test , we have :

Correct option :

H_0 : mu_D leq 0.85

H1 :11D > 0.85

(ii)

Sample mean 5%in -μD

2 0.85 ул 2/6

2.571

Correct option :

t = 2.571

(iii)

P-value can be obtained by using Excel function :  TDIST(X, n - 1= degree of freedom , 1 = one tailed test )

Pialue = TDIST(2.571, 5, 1) 0.02497 0.025

Correct option : Approximately 0.025

(iv)

Since P-value = 0.025 < 0.05 ( Level of significance) , we reject Ho and conclude that the preparation program will increase the ACT score by more than 0.85 points.

Correct option : Conclude H1 that the preparation program will increase the ACT score by more than 0.85 points.

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