Question

With individual lines at its various windows, a post office finds that the standard deviation for normally distributed waiting times for customers on Friday afternoon is 7.2 minutes. To test this claim, the post office experiments with a single, main waiting line and finds that for a random sample of 25 customers, the waiting times for customers have a standard deviation of 3.5 minutes.With a significance level of 5 percent, test the claim that a single line causes lower variation among waiting times (shorter waiting times) for customers. Degrees of freedom test statistic pvalue accept or reject

Answer 1

Chi Square statistic is 5.67129 is in the rejection region as the Chi Square critical is 13.84 842 we reject the null hypothesis σ square =7.2 square

Explanation:

calculate the chi square stat as n-1 x S square/σ square

n= 25 S=3.5 and alpha=.05 . Look at the chi square table df = 24 and area =.95 to get chi-square critical for alpha = .05 this value is 13.84842. the calculated chi square is in the rejion of rejection as this is a left tailed test.

We reject the null hypothesis . p- value need not be found but can be easily done using a calculator

Answer 2

We reject H₀ we found that at 5% significance level the single file causes shorther waiting times

Explanation:

We will develop a chi-square test since it is a standard deviation test.

Conditions for chi-square test

1.-Population follows a normal distribution

2.-We have a random sample

Therefore

Test Hypothesis

Null hypothesis H₀ σ = 7,2

Alternative hypothesis Hₐ σ < 7,2

So it is a one tailed-test to the left

Sample size n = 25

Then df = 25 - 1 df = 24

Significance level α = 5% α = 0,05

With df = 24 and α = 0,05 we find

Χ₀ = 36,415

Computing X(s)

X(s) = ( n - 1 ) * σ² / (7,2)²

X(s) = 24* (3,5)² / 51,84

X(s) = 294/51,84

X(s) = 5,67

Comparing Χ₀ and Χ(s)

X(s) < Χ₀

So X(s) is in the rejection region and we reject H₀