Question

Historically, the time needed for college students to complete their degree follows a normal distribution with a mean of 4 years and a standard deviation of 1.2 years. You wish to see if the mean time m has changed in recent years, so you collect information from 5 recent college graduates.

                                                       4.25      4            3.75       4.5 5

Is there evidence that the mean time is different from 4 years?

a. Check the needed conditions for both the test statistic and confidence interval. (Do not do a stemplot.)

b. State Ho and Ha

c. Calculate the test statistic (if applicable – state the degrees of freedom)

d. Find the p-value

e. What is the conclusion for this problem? Do you reject Ho?

f. Calculate the 98% confidence interval.

g. Interpret this confidence interval

Answer 1

Solution

Let X = the time needed (in years) for college students to complete their degree. We are given

X ~ N(4, 1.2) ........................................................................................................................................................................... (1)

Part (a)

Vide (1), the base variable follows Normal distribution. That is the only condition for both the test statistic and confidence interval. So, the needed conditions are fulfilled. Answer 1

Part (b)

Hypotheses:

Null H₀: µ = µ₀ = 4.0    Vs Alternative H_A: µ ≠ 4.0 Answer 2

[Note: ‘You wish to see if the mean time m has changed in recent years’ which implies positive as well as negative change.]

Part (c)

Test statistic:

Z = (√n)(Xbar - µ₀)/σ = 0.5590 Answer 3

where

n = sample size;

Xbar = sample average;

σ = known population standard deviation.

Since the distribution of test statistic is Normal, degrees of freedom do not apply. Answer 4

Summary of Excel Calculations is given below:

n	5
Xbar	4.3
Given
µ0	4
σ	1.2
Zcal	0.559017
Given α	0.02
p-value	0.57615

Part (d)

Distribution, Level of Significance, α, and p-value

Under H₀, Z ~ N(0, 1)

p-value = P(Z > | Zcal |)

Using Excel Function: Statistical NORMSDIST, p-value is found to be 0.5762 Answer 5

Part (e)

Decision and Conclusion:

Since p-value > α. H₀ is accepted. Answer 6

Note

Significance level α is not explicitly stipulated in the question. But, given 98% confidence interval, it is meaningful to assume α = 0.02.

Part (f)

100(1 - α) % Confidence Interval for μ, when σ is known:

Xbar ± MoE

Where

MoE = (Z_{α /2})σ/√n

With

Xbar = sample mean,

Z_{α /2} = upper (α/2)% point of N(0, 1),

σ = population standard deviation and

n = sample size.

Substituting the given values, 98% Confidence Interval for μ is: [3.05, 5.54] Answer 7

Part (g)

Interpretation

There is only 1% chance that the mean time needed for college students to complete their degree would be less than 3.05 years and only 1% chance that it will be more than 5.54 years. Answer 8

Yet another interpretation is:

Since the 98% Confidence Interval does contain 4, the null hypothesis μ = 4, is accepted at 2% significance level, confirming the earlier decision. Answer 9

DONE