Question

DO NOT USE EXCEL SHOW STEPS EXPLAIN ALL PARTS

A group of 18 people from New York and a group of 15 people from Los Angeles passed the same quiz. The mean grade of group A is 78 points, with the standard deviation σ1=5, the mean grade of group B is 75 points, σ2=4.5. Use α=0.05. Assuming that the subjects are chosen randomly, the cities’ population are independent, and the points are normally distributed:

(a) Check if there is a significant difference between the mean grades of the groups. Find the P-value of this test.

(b) Explain how the test could be conducted with a confidence interval.

(c) What is the power of the test for the true difference in means of 2.5?

(d) Assume that sample sizes are equal. What group size should be considered to obtain β=0.1 if the true difference in means is 2.5?

Answer 1

a) The null and alternative hypothesis

Test statistic is

given 78, , 5 , 4.5

thus

As the alternative hypothesis is non directional , we find two tailed P value

P value = P( z <-1.81)+P(z>1.81)= 0.0351 +0.0351 = 0.0702 (from z table)

level of significance =0.05

As P value > 0.05

We fail to reject H0

There is not sufficient evidence to conclude that there is significant difference between mean grades of two groups.

b) The 95% confidence interval for difference in mean grades

For 95% confidence , zc =1.96

Therefore , 95% confidence interval for difference in mean grades

= (-0.24 , 6.24 )

Since the confidence interval include zero , there is not sufficient evidence to conclude that there is significant difference between mean grades of two groups.

Note : There is 95% chance that the Confidence interval contain the true poprlation mean difference.As zero is also within this interval , we cannot say that the true poprlation mean difference is not zero. That means there is not significant difference between mean grades of two groups.

c) Power of the test = P( reject H0 I H0 is false )

For , the critical value of z are -1.96 and 1.96

zc =-1.96

zc =1.96

Reject H0 if

Power

=

= P( z < -3.47 ) +P( z > 0.45 )

= 0.0003 + 0.3264

=0.3266

(d) Given

that is Power = 1-0.01 =0.9

Let n1=n2 =n

Reject H0 if

that is if

From z table , P( z < -0.13 )=P(z>0.13) =0.45

Thus each sample size = 24

Combined sample size = 48