Question

In: Math

Researchers sample 3000 households concerning tv shows they watch. Based on the sample 15% reported watching...

Researchers sample 3000 households concerning tv shows they watch. Based on the sample 15% reported watching 60 minutes. What is the 95% confidence interval for the proportion of all Americans that watch 60 minutes

Solutions

Expert Solution

Solution :

Given that,

n = 3000

= 15%=0.15

1 - = 1 - 0.15 = 0.85

At 95% confidence level the z is ,

= 1 - 95% = 1 - 0.95 = 0.05

/ 2 = 0.05 / 2 = 0.025

Z/2 = Z0.025 = 1.96

Margin of error = E = Z / 2 * (( * (1 - )) / n)

= 1.96 * (((0.15 * 0.85) / 3000 = 0.01278

A 95 % confidence interval for proportion p is ,

- E < P < + E

0.15 - 0.01278 < p < 0.15 + 0.01278

0.13722 < p < 0.16278

The 95% confidence interval for the proportion of all Americans that watch 60 minutes p is :

( 0.13722, 0.16278)

Sun, Jul 15, 9:40 PM

milcah answered 1 year ago
Next > < Previous

Related Solutions

Q4.(15) One random sample of the hours of one type of user watch TV per day...
Q4.(15) One random sample of the hours of one type of user watch TV per day was recorded: 2.9, 3.5, 3, 3.2, 3.3, 3.5, 3.3, 3, 3.5. 1.(5) Apply the backward empirical rule to check normality of the data, and conclude if any evidence of non-normality. 2.(5) Assume that the underlying population is normal. Find a two-sided 99% confidence interval for the the true population average hours 3.(5) Assume that population variance σ2=0.09σ2=0.09 and the bound on the error of...
A survey stated that 40% of people in Indiana watch TV shows at night. (a) At...
A survey stated that 40% of people in Indiana watch TV shows at night. (a) At α = 0.05, test this claim using a random sample of 100 Indiana residents, out of whom 20 said that they do not watch TV shows at night. (b) Would the test results be the same if 40 people instead said they watch TV shows at night? Suppose we want to compare the results with Ohio population results of the survey, where it has...
One random sample of the hours of one type of user watch TV per day was...
One random sample of the hours of one type of user watch TV per day was recorded: 2.9, 3.5, 3, 3.2, 3.3, 3.5, 3.3, 3, 3.5. 1. Apply the backward empirical rule to check normality of the data, and conclude if any evidence of non-normality. 2. Assume that the underlying population is normal. Find a two-sided 99% confidence interval for the the true population average hours. 3. Assume that population variance ?2=0.09σ2=0.09 and the bound on the error of estimation...
EXPA media research reported that adult women watch TV an average of 5.3 hrs/day, compared to...
EXPA media research reported that adult women watch TV an average of 5.3 hrs/day, compared to an average of 4.6 hrs/day for adult men. Assume that those results are found from a sample of 120 men and 160 women and that two population standard deviations are minimum 0.48 and 0.60 respectively. a) test the belief that women on average spend more time watching TV than men. Use a 0.05 significance level. H0: H1: Test statistic: b) construct a 90% confidence...
h2. The table below shows a sample of households and two variables: (1) the number of...
h2. The table below shows a sample of households and two variables: (1) the number of people in the household (2) the weekly pounds of trash generated by the household People in house Pounds of Trash per Week 2 18 3 33 6 93 1 23 7 83 Explain the relationship between the amount of trash per week and the number of people in the household using concepts from linear regression. Provide a comprehensive answer that includes explaining the following:...
​​​​​​A sample survey of 36 randomly selected households shows that the median household income of the...
​​​​​​A sample survey of 36 randomly selected households shows that the median household income of the eastern NC residents is $51,000, with a standard deviation of $1075. Test the (null) hypothesis that the actual (real) median household income is $52,752 at 0.05% confidence level.
The Food Marketing Institute shows that 15% of households spend more than $100 per week on...
The Food Marketing Institute shows that 15% of households spend more than $100 per week on groceries. Assume the population proportion is p = 0.15 and a sample of 600 households will be selected from the population. Use z-table. Calculate (), the standard error of the proportion of households spending more than $100 per week on groceries (to 4 decimals).    What is the probability that the sample proportion will be within +/- 0.03 of the population proportion (to 4...
The Food Marketing Institute shows that 15% of households spend more than $100 per week on...
The Food Marketing Institute shows that 15% of households spend more than $100 per week on groceries. Assume the population proportion is p = 0.15 and a sample of 800 households will be selected from the population. Use z-table. 1. Calculate (), the standard error of the proportion of households spending more than $100 per week on groceries (to 4 decimals). 2. What is the probability that the sample proportion will be within +/- 0.02 of the population proportion (to...
The Food Marketing Institute shows that 15% of households spend more than $100 per week on...
The Food Marketing Institute shows that 15% of households spend more than $100 per week on groceries. Assume the population proportion is p = 0.15 and a sample of 800 households will be selected from the population. A) What is the probability that the sample proportion will be within +/- 0.03 of the population proportion (to 4 decimals)? B) What is the probability that the sample proportion will be within +/- 0.03 of the population proportion for a sample of...
A sample of 250 apartment house shows that 154 house subscribes to cable TV. Construct a...
A sample of 250 apartment house shows that 154 house subscribes to cable TV. Construct a 90% confidence interval for the proportion of all apartment houses that subscribes to cable TV. Also, find the margin of error. Round your answers to 3 decimal places.
ADVERTISEMENT
Subjects
ADVERTISEMENT
Latest Questions
ADVERTISEMENT