Respuesta :
Answer: b. 0.98
Step-by-step explanation:
The formula to find the maximum error of the estimated mean :
[tex]E=z^*\dfrac{\sigma}{\sqrt{n}}[/tex] (1)
, where [tex]\sigma[/tex] = standard deviation
n= Sample size
z* = Critical z-value.
As per given , we have
[tex]\sigma=5[/tex]
n=100
Critical value for 95% confidence interval = z*=1.96
Put these values in the formula (1), we get
[tex]E=(1.96)\dfrac{5}{\sqrt{100}}[/tex]
[tex]E=(1.96)\dfrac{5}{10}=0.98[/tex]
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:
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, [tex]\sigma[/tex] = 5
Sample size, n = 100
Now, the Margin of error formula is given by = [tex]Z_\frac{\alpha}{2} *\frac{\sigma}{\sqrt{n} }[/tex]
Here, [tex]Z_\frac{\alpha}{2}[/tex] = value of z score at 2.5% significance level which is = 1.96 {using z table}
So, Margin of error = [tex]1.96 *\frac{5}{\sqrt{100} }[/tex]
= 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 .
