Question

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 95% level of confidence and an estimated standard deviation of 5?

a. 0.9604

b. 0.98

c. 1.96

d. 5%

Answer 1

Option b = 0.98

Explanation:

We are given that a local company wants to evaluate their quality of service by surveying their customers. Their budget limits the number of surveys to 100.

Also, Confidence level = 95%

Standard deviation,
`\sigma` = 5

Sample size, n = 100

Now, the Margin of error formula is given by =
`Z_(\alpha)/(2) *(\sigma)/(√(n) )`

Here,
`Z_(\alpha)/(2)` = value of z score at 2.5% significance level which is = 1.96 {using z table}

So, Margin of error =
`1.96 *(5)/(√(100) )`

= 1.96 * 0.5 = 0.98

Therefore, maximum error of the estimated mean quality for a 95% level of confidence and an estimated standard deviation of 5 is 0.98 .

Answer 2

Answer: b. 0.98

Explanation:

The formula to find the maximum error of the estimated mean :

`E=z^*(\sigma)/(√(n))` (1)

, where
`\sigma` = standard deviation

n= Sample size

z* = Critical z-value.

As per given , we have

`\sigma=5`

n=100

Critical value for 95% confidence interval = z*=1.96

Put these values in the formula (1), we get

`E=(1.96)(5)/(√(100))`

`E=(1.96)(5)/(10)=0.98`

Hence, the maximum error of the estimated mean quality for a 95% level of confidence is 0.98.

Therefore , the correct answer is b. 0.98 .