Question

1. You are a teacher at a school for gifted children, and you feel that your newest class of students are even brighter than usual. The mean IQ at your school is 127, and the mean IQ of this new class is 134. In total, there are 32 students in this new class. Also, the standard deviation of the school's IQ is 8. Use the eight steps to test whether this new class of students is significantly more intelligent than the school's student body overall. What is the effect size between the sample and population? In the population the average IQ is 100 with a standard deviation of 15. A team of scientists want to test a new medication to see if it has either a positive or negative effect on 10, or no effect at all. A sample of 30 participants who have taken the medication have a mean of 140. Did the medication affect intelligence? What is the effect size between the sample and population of question 107 Interpret the following result: 2 = 3.49. p <.05 (give what Z represents, 3.49 represents and what p<.05 represents

Answer 1

STEP 1)NULL HYPOTHESIS H0:

μ = 127

STEP2) ALTERNATIVE HYPOTHESIS Ha:

μ2 127

This corresponds to a right-tailed test, for which a z-test for one mean, with known population standard deviation will be used.

(STEP 3) Rejection Region: Based on the information provided, the significance level is α=0.05, and the critical value for a right-tailed test is zc​=1.64.

The rejection region for this right-tailed test is R={z:z>1.64}

(STEP 4) Test Statistics

The z-statistic is computed as follows:

2-1-Ho_134-127 20/ = 4.95 8/32 :

STEP 5 : DEGREES OF FREEDOM= N-1=32-1=31

(STEP 6) Decision about the null hypothesis: Since it is observed that z=4.95>zc​=1.64, it is then concluded that the null hypothesis is rejected.

STEP 7: Using the P-value approach: The p-value is p=0, and since p=0<0.05, it is concluded that the null hypothesis is rejected.

(STEP 8) Conclusion: It is concluded that the null hypothesis Ho is rejected. Therefore, there is enough evidence to claim that the population mean μ is greater than 127, at the 0.05 significance level.

EFFECT SIZE=

Χ– μισ

=

134-127/8=

= 0.875

Effect size is large= 0.875

SOLUTION 2: (1) Null and Alternative Hypotheses

The following null and alternative hypotheses need to be tested:

Ho: μ=100

Ha: μ≠​100

This corresponds to a two-tailed test, for which a z-test for one mean, with known population standard deviation will be used.

(2) Rejection Region

Based on the information provided, the significance level is \alpha = 0.05α=0.05, and the critical value for a two-tailed test is zc​=1.96.

The rejection region for this two-tailed test is R={z:∣z∣>1.96}

(3) Test Statistics

The z-statistic is computed as follows:

- * Χ - μo σι η 140 - 100 15/4/30

(4) Decision about the null hypothesis: Since it is observed that ∣z∣=14.606>zc​=1.96, it is then concluded that the null hypothesis is rejected.

Using the P-value approach: The p-value is p=0.000, and since p=0.000<0.05, it is concluded that the null hypothesis is rejected.

(5) Conclusion: It is concluded that the null hypothesis Ho is rejected. Therefore, there is enough evidence to claim that the population mean μ is different than 100, at the 0.05 significance level. Yes the medication does effect the intelligence.

Confidence Interval: The 95% confidence interval is 134.632<μ<145.368.

Effect Size=

Τ– μσ = 140 – 100/15 = 40/15 = 2.67

EFFECT SIZE IS VERY LARGE

Z=3.49 IS VALUE OF TEST STATISTIC AND P VALUE SMALLER THAN 0.05 WHICH MEANS TEST STATISTIC IS SIGNIFICANT. Here P value smaller so there is stronger evidence in favor of the alternative hypothesis..