Question

e Section 3: 2 Scores/Percentiles and Hypothesis Testing Please show work. You wil need to refer to Z score tables for this rtin 6. The national (population) average for ACT scores is 208 (on a 36 point a. Suppose you score 26 on does this score fall? (Find your ACT. At approximately what percentile (round to the nearest integer between 1 and 99) your probability from the Z table, subtract that from 1, multiply by 100, and round) Answer that instead you have sample of this population, consisting of 400 students, all coming from the same school heir average ACT score is 22.1. One distressed student among them scored 19 and would like to know his percentile amongst his peers. Applying the same process as in part 6a. (but with some minor differences; see hint below), find this student's percentile. (Hint: s is used to indicate the estimated (sample) standard deviation, given a sample of size N from a population with a known standard deviation of Ơ They are related: s-aWN, in this case, wings or sould dramatically affect the normalization. Note above that this explicitly says sample of this population) Answer e. The principal of this school claims that his students perform better than the national average on the ACT set up a hypothesis test to (with a significance of 10%, or 0.1) to determine whether this principars claim is valid, and justify t with Z scores (alpha) and p values Standard Normal Distribution 10 meen, unit voriance, right-tail If p is here, reject H 02

Answer 1

6.
a.

Z score for score 26 = (x - $\mu$) / $\sigma$ = (26 - 20.8) / 4.8 = 1.083

Percentile for Z = 1.083 is 0.86 = 86% (From Z table)

b.

Sample average = 22.1

Z score for score 19 = (19 - 22.1) / 4.8 = -0.6458

Percentile for Z = -0.6458 is 0.2592 = 26% (From Z table)

c.

Null hypothesis H0: Mean score of students is equal to the national average on ACT. That is, $\mu$ = 20.8
Alternative hypothesis Ha: Mean score of students is greater than the national average on ACT. That is, $\mu$ > 20.8

Standard error of mean = s / $\sqrt{n}$ = 4.8 / $\sqrt{400}$ = 0.24

Test statistic, z = (Sample mean - Hypothesized mean) / Standard error

= (22.1 - 20.8) / 0.24

= 5.42

Critical value of Z for $\alpha$ = 0.10 and one-tail test is 1.28. That is, we reject H0 if Z > 1.28

Since test statistic, z is greater than the critical value, we reject H0.

P-value = P(Z > 5.42) = 0.0000

Since, p-value is less than 0.1 significance level, we reject null hypothesis H0 and conclude that there is significant evidence that mean score of students is greater than the national average on ACT and the principal's claim is valid.