Question

g The manager of a grocery store has selected a random sample of 100 customers. The average length of time it took these 100 customers to check out was 3.0 minutes. It is known that the standard deviation of the checkout time is 1 minute. Refer to Exhibit 8-2. The 95% confidence interval for the average checkout time for all customers is _____. a. 2.804 to 3.196 b. 1.04 to 4.96 c. 1.36 to 4.64 d. 3 to 5

Answer 1

Answer:

The right solution is "0.196".

Step-by-step explanation:

The given values are:

Standard deviation,

$\sigma=1 \ minute$

Sample size,

$n=100$

Confidence level,

$c=95$%

  $=0.95$

Now,

For $\alpha=1-c$

         $= 1-0.95$

         $=0.05$

then,

$Z_{\frac{\alpha}{2} } = Z_{\frac{0.05}{2} }$

     $=Z_{0.025}$

     $=1.96$

The margin or error will be:

⇒ $E=Z_{\frac{\alpha}{2} } (\frac{\sigma}{\sqrt{n} })$

On substituting the values, we get

       = $1.96(\frac{1}{\sqrt{100} })$

       = $1.96\times \frac{1}{10}$

       = $0.196$