A researcher examined a random sample of 300 homes in a small city and found that 52 had solar panels installed on their roofs. Use the sample to construct a 99% confidence interval for the proportion of all homes in the city that have solar panels installed on their roofs.
What is the 99% confidence interval?
We have given here,
x=52
n=300
Estimate for sample proportion
Level of significance is =1-0.99=0.01
Z critical value(using Z table)=2.576
A researcher examined a random sample of 300 homes in a small city and found that...
A researcher examined a random sample of 300 homes in a small city and found that 52 had solar panels installed on their roofs. Use the sample to construct a 99% confidence interval for the proportion of all homes in the city that have solar panels installed on their roofs. What is the margin of error for the 99% confidence interval? Round to three decimal places.
A researcher examined a random sample of 300 homes in a small city and found that 52 had solar panels installed on their roofs. Use the sample to construct a 99% confidence interval for the proportion of all homes in the city that have solar panels installed on their roofs. What is the margin of error for the 99% confidence interval? Round to three decimal places.
A researcher examined a random sample of 150 homes in a small city and found that 24 had solar panels installed on their roofs. Use the sample to construct a 99% confidence interval for the proportion of all homes in the city that have solar panels installed on their roofs. What is the sample proportion? Round to two decimal places.
In a random sample of 100 homes in a certain city, it is found that 10 are heated by oil. (a) Find a 92% confidence interval for the true proportion of homes in the city that are heated with oil. (b) How large would the sample have to be in order to have a standard error that is 0.003 or lower?
A random sample of 366 married couples found that 298 had two or more personality preferences in common. In another random sample of 574 married couples, it was found that only 22 had no preferences in common. Let p1 be the population proportion of all married couples who have two or more personality preferences in common. Let p2 be the population proportion of all married couples who have no personality preferences in common. (a) Find a 99% confidence interval for...
A random sample of 388 married couples found that 280 had two or more personality preferences in common. In another random sample of 562 married couples, it was found that only 36 had no preferences in common. Let p1 be the population proportion of all married couples who have two or more personality preferences in common. Let p2 be the population proportion of all married couples who have no personality preferences in common. (a) Find a 99% confidence interval for...
A. You selected a simple random sample of 225 households from a city. In this sample of 225 households, 81 households subscribe to Netflix. Let π denote the proportion of households in this city who subscribe to Netflix. Construct a 99% confidence interval for π. (Show work) B. A city has 50,000 households. You have drawn a simple random sample of 400 households from this city. In this sample of 400 households, 60 households own hybrid vehicles. At a 99%...
1) From a random sample of 121 government employees it is found that 65 of them are males. Construct a 99% confidence interval for the proportion of males in government service.
Based on a random sample of 16 trees, a forestry researcher found that the average height was 4.8 meters with a standard deviation of 0.55 meters. Assume that the distribution of the heights of these trees is normally distributed and does not have any outliers, and that the appropriate multiplier for 90% confidence in 1.74. Use this information to calculate the quantities requested below What is the standard error of y? What is the margin of error for the 90%...
A random sample of 300 items reveals that 192 of the items have the attribute of interest. a. What is the point estimate for the population proportion for all items having this attribute? b. Use the information from the random sample to develop a 90% confidence interval estimate for the population proportion, p, of all items having this attribute of interest.