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. a) With a significance level of 5 percent, conduct an appropriate test on the claim. (State your null hypothesis, test statistic, critical value, decision rule, conclusion) (15 points) b) What is the 95% confidence interval for the population variance

Answer 1

Answer:

H0 : σ²= 7.2²

H1 : σ² < 7.2

We can conclude that the mean waiting time of customers vary less, Than 7.2 minutes.

Step-by-step explanation:

H0 : σ²= 7.2²

H1 : σ² < 7.2²

The test statistic :

X² = [(n - 1)*s² ÷ σ²]

s² = 3.5

n = sample size = 25. ; s²

X² = [(25 -1)*3.5² ÷ 7.2²

X² = (24 * 3.5^2) / 7.2^2

X² = 5.67

Pvalue from Chisquare statistic :

P(X² < 5.67) = 0.000042

Pvalue < α ; we reject the Null.

Hence, we can conclude that the mean waiting time of customers vary less, Than 7.2 minutes.