A local company wants to evaluate their quality of service by surveying their customers. Their budget limits the number of surveys to 100. What is their maximum error of the estimated mean quality for a 93% level of confidence and an estimated standard deviation of 7? a. 1.8130 b. 1.1550 c. 1.2824 d. 1.3160

(C) 1.2824

Step-by-step explanation:

Maximum error = t * sd/√n

sd = 7

n = 1

degree of freedom = n - 1 = 100 - 1 = 99

confidence level = 93%

t-value corresponding to 99 degrees of freedom and 93% confidence level is 1.855

Maximum error = 1.855*7/√100 = 12.985/10 = 1.2985

The closest option is (C)

Answer:1.2684

Step-by-step explanation:

Formula to find the maximum error of the mean is given by : -

E=z*/dfrac{/sigma}{/sqrt{n}}

, where n = sample size.

z * = Critical value.

= Population standard deviation

As it is given, we have

n = 100

/sigma = 7

93%: Confidence interval

Critical value for 93% confidence as gotten from the z-table is = 1.81 [from z-table ]

Then, estimated mean quality has the maximum error as

E = (1.81) / dfrac{7}{/sqrt{100}}

E = (1.81) / dfrac{7}{10}

E = (1.81) / dfrac07=1.2684

Therefore, maximum error required will be = 1.2684