Question

The Jefferson Valley Bank once had a separate customer waiting line at each teller window, but it now has a single waiting line that feeds the teller windows as vacancies occur. The standard deviation of customer waiting times with the old multiple-line configuration was 1.8 min. Listed below is a simple random sample of waiting times (minutes) with the single waiting line. Use a 0.05 significance level to test the claim that with a single waiting line, the waiting times have a standard deviation less than 1.8 min. What improvement occurred when banks changed from multiple waiting lines to a single waiting line?

Answer 1

Answer:

There will be sufficient evidence to conclude. A further explanation is provided below.

Step-by-step explanation:

The given values are:

$\sigma=1.8$

$\alpha=0.05$

$n=10$

As we know,

$\bar{x}=\frac{\Sigma x_i}{n}$

  $=\frac{71.5}{10}$

  $=7.15$

The standard deviation will be:

⇒  $s=\sqrt{\frac{1}{n-1} \Sigma(x_i- \bar{x})^2 }$

On substituting the values, we get

⇒     $=\sqrt{\frac{1}{10-1} [(6.5-7.15)^2+...(7.7-7.15)^2]}$

⇒     $=\sqrt{\frac{1}{9} [(6.5-7.15)^2+...(7.7-7.15)^2]}$

⇒     $=0.477$

According to the question,

Hypotheses:

$H_o: \sigma=1.8$

$H_a: \sigma$

The test statistic will be:

⇒ $X^2=\frac{(n-1)s^2}{\sigma^2}$

⇒       $=\frac{(10-1)\times 0.477^2}{1.8^2}$

⇒       $=\frac{2.0477}{3.24}$

⇒       $=0.632$

Thus the above is the correct response.