The manager of a grocery store has taken a random sample of

The manager of a grocery store has taken a random sample of n 50 customers. 17k average length ot time it took these 50 customers to check out was x =3.5 minutes. It is also tound that the standard deviation of these 50 customers to check out was -v 1.25 minutes. Use the appropriate method to answer the following questions. No marks will he awarded if you use incorrect method. Construct a 90% confidence interval for mu-the mean checkout time (in minutes) of all customers. Interpret this interval. Show your complete work and w rite the formula clearly. Construct a 80% confidence interval for ft =the mean checkout time (in minutes) of all customers. Interpret this interval. Show your complete w ork and write the formula clearly. What do you think about the and the confidence interval in parrs (a)-(b) as the confidence level is increased? Explain clearly.

Solution

a)
Confidence Interval
CI = x ± t a/2 * (sd/ Sqrt(n))
Where,
x = Mean
sd = Standard Deviation
a = 1 - (Confidence Level/100)
ta/2 = t-table value
CI = Confidence Interval
Mean(x)=3.5
Standard deviation( sd )=1.25
Sample Size(n)=50
Confidence Interval = [ 3.5 ± t a/2 ( 1.25/ Sqrt ( 50) ) ]
= [ 3.5 - 1.677 * (0.177) , 3.5 + 1.677 * (0.177) ]
= [ 3.204,3.796 ]

b.
At ALpha - 80%
Confidence Interval = [ 3.5 ± t a/2 ( 1.25/ Sqrt ( 50) ) ]
= [ 3.5 - 1.299 * (0.177) , 3.5 + 1.299 * (0.177) ]
= [ 3.27,3.73 ]

c.
Decreasing the confidence decrement the margin of error , as well as decrease the C.I