A random sample of high school students found that 256 of the students had TVs in their rooms and that the remaining 769 did not have TVs in their rooms. Construct and interpret a 90% confidence interval for the proportion of high school students who have TVs in their rooms.

1. Show that this situation satisfies the following conditions:
a. The 10% condition.
b. The large count condition.

2. The general formula used to create a confidence interval is:

Statistic ± (critical value) × (standard deviation of statistic)

a. What is the statistic (or point estimate) in this example?
b. What is the critical value (rounded to 4 decimal places) to determine the confidence interval?
c. What is the standard deviation of the statistic?

The conditions for constructing a 90% confidence interval for the proportion of high school students with TVs in their rooms were verified. A point estimate of 0.25 was found. Using this, a standard deviation of the statistic was computed, and the confidence interval was created.

It is Statistics which deals with data interpretation and probability. We'll make a 90% confidence interval for the proportion of high school students who have TVs in their rooms. There are two conditions to meet: the 10% condition, and the large count condition.

Firstly, for sampling to be performed without replacement, the sample size must be no greater than 10% of the population. This is known as the 10% condition. With no information about the total number of high school students, we assume it's larger enough that 256 is less than 10% of the population.

Next, the large count condition requires that we have at least 10 successes and 10 failures. Here, numbers of students with TVs (256) and without TVs (769) are both larger than 10, so large count condition is satisfied.

The point estimate (or statistic) is the proportion of students with TVs, 256/(256+769) = 0.25 (rounded to two decimal places). For a 90% confidence level, the critical value (z-star) is about 1.645. The standard deviation of the proportion (sqrt[p(1-p)/n]) can be calculated using the point estimate from step one, approximately 0.014. We plug these into the confidence interval formula to construct the interval.

We can be 90% confident that the actual proportion of high school students with TVs in their rooms lies between the calculated interval.

To know more about standard deviation visit:

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