Asample of 412 adults showed that 293 of them are connected the the internet from home. construct and interpret a 90% confidence interval for the proportion of all adults that have internet access.

Given:

Sample no. of events,

Sample size,

Now,

The sample proportion will be:

→

The significance level will be:

Form the z-table,

The critical value,

Now,

The standard error will be:

=

=

and,

The margin of error,

→

Now,

The lower limit will be:

=

=

The upper limit will be:

=

=

hence,

The CI is "(0.6744, 0.748)". Thus the response above is right.

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CI = (0.674, 0.748)

Step-by-step explanation:

The confidence interval of a proportion is:

CI = p ± SE × CV,

where p is the proportion, SE is the standard error, and CV is the critical value (either a t-score or a z-score).

We already know the proportion:

p = 293/412

p = 0.711

But we need to find the standard error and the critical value.

The standard error is:

SE = √(p (1 − p) / n)

SE = √(0.711 × (1 − 0.711) / 412)

SE = 0.0223

To find the critical value, we must first find the alpha level and the degrees of freedom.

The alpha level for a 90% confidence interval is:

α = (1 − 0.90) / 2 = 0.05

The degrees of freedom is one less than the sample size:

df = 412 − 1 = 411

Since df > 30, we can approximate this with a normal distribution.

If we look up the alpha level in a z score table or with a calculator, we find the z-score is 1.645.  That's our critical value.  CV = 1.645.

Now we can find the confidence interval:

CI = 0.711 ± 0.0223 * 1.645

CI = 0.711 ± 0.0367

CI = (0.674, 0.748)

So we are 90% confident that the proportion of adults connected to the internet from home is between 0.674 and 0.748.

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step-by-step explanation: