# MGEB12H3 Lecture Notes - Lecture 1: Null Hypothesis, Confidence Interval, Test StatisticExam

by OC2447826

Department

Economics for Management StudiesCourse Code

MGEB12H3Professor

ATA Lecture

1This

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TUTORIAL-1

Long Question: The manager of a department store is thinking about establishing a new

billing system for the store's customers. After a thorough financial analysis, she

determines that the new system will be cost-effective only if the mean monthly account

balance is greater than $250. A random sample of 111 monthly accounts is drawn. The

sample mean is $272.50, and the sample standard deviation is $130.00.

a) Identify the Null and Alternative hypotheses for a hypothesis test that would allow

the manager to decide whether there is sufficient evidence to conclude that the new

system will be cost-effective. Using a value of α = .05, identify the critical value(s) of

the test statistic used for this hypothesis test. State your conclusion.

Solution: H0:

= 250 H1:

> 250

Answer: t.05, 110 = 1.6588

Test Statistic: t =

8235.1

111

130

00.25050.272

0

n

S

X

Decision: Reject the null hypothesis.

Conclusion: There is enough evidence to indicate that the new system is cost effective.

b) Using the t-table show how you can approximate the p-value of your test. in (a).

Solution:

0.025<p-value <0.05

the t-stat of 1.8235 is between 1.658, 1.980 from the table.

c) Calculate the 95% confidence interval for the population mean. Briefly explain the

meaning of the confidence interval. Based on this confidence interval redo your test

in (a).

Solution:

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43

.

24

50

.

272

111

130

98.150.272

2/

n

s

tX

The null as appeared in (a) is inside the confidence interval implying that the

null cannot be rejected at 2.5%, (Note: this is a one sided test)). We cannot

reject at 2.5% but that does not imply we cannot reject at 5%. Therefore based

on the confidence interval alone we cannot do the test in (a).

d) Find the smallest value for

X

that allows you to reject the null in (a).

Answer: To reject the null hypothesis, we would need to see a test statistic that is more

extreme than the critical value of 1.6588:

Test Statistic: t =

111

130

00.250X

= 1.6588;

X

= 270.47

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