Question

A local company wants to evaluate their quality of service by surveying their customers.

Answer 1

Answer: b. 0.98

Step-by-step explanation:

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

$E=z^*\dfrac{\sigma}{\sqrt{n}}$            (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

$E=(1.96)\dfrac{5}{\sqrt{100}}$

$E=(1.96)\dfrac{5}{10}=0.98$

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 .

Answer 2

Answer:

Option b = 0.98

Step-by-step 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 = $Z_\frac{\alpha}{2} *\frac{\sigma}{\sqrt{n} }$

Here, $Z_\frac{\alpha}{2}$ = value of z score at 2.5% significance level which is = 1.96 {using z table}

So, Margin of error = $1.96 *\frac{5}{\sqrt{100} }$

                               = 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 .