Question

A random sample of 300 CitiBank VISA cardholder accounts indicated a sample mean debt of 1,220 with a sample standard deviation of 840. Construct a 95 percent confidence interval estimate of the average debt of all cardholders.

Answer 1

Answer: (1124.5619, 1315.4381)

Step-by-step explanation:

The confidence interval for population mean$(\mu)$ when populatin standard deviation is unknown :-

$\overline{x}\pm t^*\dfrac{s}{\sqrt{n}}$

, where $\overline{x}$ = Sample mean

$s$ =Sample standard deviation

t* = Critical t-value.

Given : n= 300

Degree of freedom : df = n-1 = 299

$\overline{x}=1220$

$s=840$

Confidence interval = 95%

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

Using t-distribution table ,

The critical value for 95% Confidence interval for significance level 0.05 and df = 299 : $t^*=t_{\alpha/2,\ df}=t_{0.025,\ 299}=1.9679$

Then, a 95% confidence interval estimate of the average debt of all cardholders will be :-

$1220\pm (1.9679)\dfrac{840}{\sqrt{300}}$

$=1220\pm (1.9679)\dfrac{840}{17.3205080757}$

$=1220\pm (1.9679)(48.4974226119)$

$\approx1220\pm 95.4381=(1220-95.438,\ 1220+95.438)\\\\=(1124.5619,\ 1315.4381)$

Hence, a 95% confidence interval estimate of the average debt of all cardholders is (1124.5619, 1315.4381) .