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Nookie1986 [14]
3 years ago
5

N a simple random sample of 219 students at a college, 73 reported that they have at least $1000 of credit card debt.

Mathematics
1 answer:
Ivahew [28]3 years ago
3 0

Answer:

0.333 - 2.58 \sqrt{\frac{0.333(1-0.333)}{219}}=0.251

0.333 + 2.58 \sqrt{\frac{0.333(1-0.333)}{219}}=0.416

The 99% confidence interval would be given (0.251;0.416).

(25.0 ,41.6)

Step-by-step explanation:

1) Data given and notation  

n=219 represent the random sample taken    

X=73 represent the students that reported that they have at least $1000 of credit card debt.

\hat p=\frac{73}{219}=0.333 estimated proportion of students that reported that they have at least $1000 of credit card debt.

\alpha=0.01 represent the significance level

z would represent the statistic (variable of interest)    

p= population proportion of students that reported that they have at least $1000 of credit card debt.

2) Confidence interval

The confidence interval would be given by this formula

\hat p \pm z_{\alpha/2} \sqrt{\frac{\hat p(1-\hat p)}{n}}

For the 99% confidence interval the value of \alpha=1-0.99=0.01 and \alpha/2=0.005, with that value we can find the quantile required for the interval in the normal standard distribution.

z_{\alpha/2}=2.58

And replacing into the confidence interval formula we got:

0.333 - 2.58 \sqrt{\frac{0.333(1-0.333)}{219}}=0.251

0.333 + 2.58 \sqrt{\frac{0.333(1-0.333)}{219}}=0.416

And the 99% confidence interval would be given (0.251;0.416).

We are confident that about 25.1% to 41.6% of students have at least $1000 of credit card debt.

And for this case the most accurate option is:

(25.0 ,41.6)

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