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In a random sample of 145 people, 112 said that they watched educational TV. Find the 93% confidence interval of the true proportion of people who watched educational TV.

User Ricoter
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Final answer:

The 93% confidence interval for the proportion of people who watch educational TV is calculated using the sample proportion and the z-value associated with a 93% confidence level, by inputting these into the confidence interval formula for proportion.

Step-by-step explanation:

To find the 93% confidence interval for the proportion of people who watch educational TV, we first calculate the sample proportion (p-hat) by dividing the number of people who said they watched educational TV by the total number of people surveyed:

p-hat = 112 / 145

We can then use the formula for the confidence interval of a proportion, which is:

Confidence Interval = p-hat ± (Z* * sqrt((p-hat*(1-p-hat))/n))

The z-value for a 93% confidence level (Z*) can be found using a z-table or a statistical calculator. The formula sqrt((p-hat*(1-p-hat))/n) is the standard error of the proportion. After computing these values, we simply plug them into the confidence interval formula to calculate the interval.

User Yiorgos
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