Respuesta :
Answer: ($8349.69, $10274.31)
Step-by-step explanation:
When population standard deviation is unknown, then the formula to find the confidence interval for population mean is given by :-
[tex]\overline{x}\pm t_{\alpha/2}\dfrac{s}{\sqrt{n}}[/tex]
, where n= sample size
s= sample standard deviation.
[tex]\overline{x}[/tex] = sample mean
As per given , we have
Significance level : [tex]\alpha=1-0.95=0.05[/tex]
Degree of freedom : df = 69
Using t-distribution ,
Critical value for confidence interval : [tex]t_{\alpha/2,df}=t_{0.025,69}=1.9949[/tex]
[tex]\overline{x}=\$9312[/tex]
s= $4007
Now, required confidence interval for the mean credit card balance for its customers will be :-
[tex]9312\pm (1.9949)\dfrac{4007}{\sqrt{69}}[/tex]
[tex]9312\pm (1.9949)(482.3861)[/tex]
[tex]9312\pm (962.31)[/tex]
[tex](9312-962.31,\ 9312+962.31 )=(8349.69,\ 10274.31)[/tex]
The 95% confidence interval for the mean credit card balance for its customers= ($8349.69, $10274.31)
Answer:
Step-by-step explanation:
step explanation:
Assuming the credit card balance follows normal distribution,
We would determine a 95% confidence interval for the mean credit card balance.
Number of samples. n = 70
Mean, u = 9312
Standard deviation, s = 4007
For a confidence level of 95%, the corresponding z value is 1.96. This is determined from the normal distribution table.
We will apply the formula
Confidence interval
= mean ± z ×standard deviation/√n
It becomes
9312 ± 1.96 × 4007/√70
= 9312 ± 1.96 × 478.93
= 9312 ± 938.76
The lower end of the confidence interval is 9312 - 938.76 = 8373.24
The upper end of the confidence interval is 9312 - 938.76 =10250.76
Therefore, with 95% confidence interval, the mean credit card balance is between $8373.24 and $10250.76
