Question

A band is interested in determining the mean credit card balance for its customers. From a sample of 70 customers, the bank determine: X bar = $9312, sample standard deviation = $4007 Calculate the 95% confidence interval.

Answer 1

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

Answer 2

Answer: ($8349.69, $10274.31)

Explanation:

When population standard deviation is unknown, then the formula to find the confidence interval for population mean is given by :-

`\overline{x}\pm t_(\alpha/2)(s)/(√(n))`

, where n= sample size

s= sample standard deviation.

`\overline{x}` = sample mean

As per given , we have

Significance level :
`\alpha=1-0.95=0.05`

Degree of freedom : df = 69

Using t-distribution ,

Critical value for confidence interval :
`t_(\alpha/2,df)=t_(0.025,69)=1.9949`

`\overline{x}=\$9312`

s= $4007

Now, required confidence interval for the mean credit card balance for its customers will be :-

`9312\pm (1.9949)(4007)/(√(69))`

`9312\pm (1.9949)(482.3861)`

`9312\pm (962.31)`

`(9312-962.31,\ 9312+962.31 )=(8349.69,\ 10274.31)`

The 95% confidence interval for the mean credit card balance for its customers= ($8349.69, $10274.31)