Question

In: Statistics and Probability

If the population for the number of hours spend on reading a book each month is...

If the population for the number of hours spend on reading a book each month is normally distributed. A random sample of 16 students in the reading club were selected and found that, they spent an average of 70 hours each month with standard deviation of 20 hours. Construct a 95% confidence interval estimate for the population mean hours they spend on reading books each month.

Solutions

Expert Solution

Solution :

Given that,

= 70

s =20

n = 16

Degrees of freedom = df = n - 1 =16 - 1 = 15

At 95% confidence level the t is ,

= 1 - 95% = 1 - 0.95 = 0.05

  / 2= 0.05 / 2 = 0.025

t /2,df = t0.025,15 =2.131 ( using student t table)

Margin of error = E = t/2,df * (s /n)

= 2.131* ( 20/ 16)

= 10.6550

The 95% confidence interval estimate of the population mean is,

- E < < + E

70 - 10.6550 < < 70+10.6550

59.3450 < < 80.6550

( 59.3450 ,80.6550)


Related Solutions

A scientist is interested in the relationship between the number of hours that bears spend sleeping...
A scientist is interested in the relationship between the number of hours that bears spend sleeping in the week before they wake up, and the food they received on that hunt. Data was collected for four bears. Hours of sleep Food received 1 76 2 85 3 64 5 70 You are to perform a statistical analysis for the scientist. a. Note which variable is the independent, and which is the dependent variable. b. Compute the least squares estimated regression...
The variable to be measured is the number of hours a ten year old spend watching...
The variable to be measured is the number of hours a ten year old spend watching videos . Using three distributions- 1 Distribution of a population of an individual 2 Distribution of a particular sample of individuals from the population 3 Distribution of the mean Explain the difference in these three distribution to some one who has never taken statistics class before.
The variable being measured is the number of hours a 10 year old spend watching videos....
The variable being measured is the number of hours a 10 year old spend watching videos. Using three distribution- 1 Distribution of a population of individuals 2 Distribution of a particular sample of individuals fro the population 3 Distribution of mean Explain the difference in all three distributions to someone who has never taken statistics class before.
The number of hours per week that high school seniors spend on computers is normally distributed...
The number of hours per week that high school seniors spend on computers is normally distributed with a mean of 5 hours and a standard deviation of 2 hours. 70 students are chose at random, let x̅ represent the mean number of hours spent on a computer for this group. Find the probability that x̅ is between 5.1 and 5.7.
2. You are interested in the mean number of hours that Americans consume media (watching, reading,...
2. You are interested in the mean number of hours that Americans consume media (watching, reading, listening to content produced by someone else) each day. A random sample of 200 Americans consumes an average of 10.3 hours of media per day, with a standard deviation of 3.5 hours. Construct and interpret a 95% confidence interval for the true mean number of hours that Americans consume media each day.
Suppose data about the average number of hours people spend on social media websites at work...
Suppose data about the average number of hours people spend on social media websites at work is randomly collected for 6 days: 5, 4, 3.1, 2, 4.5, 1.7 A.) If we assume this population is normally distributed, construct an 99% confidence interval for the population average amount of time people spend on social media websites at work. B.) Suppose I want to know more about how workers spend their time throughout the day. I want to test the claim that...
For the population of Cal Poly students, let X = number of hours slept in the...
For the population of Cal Poly students, let X = number of hours slept in the last 24 hours and Y = number of exams today. Consider the following (admittedly simplistic) joint probability distribution for Xand Y. x 5 6 7 8 0 .01 .09 .16 .18 y 1 .11 .06 .04 .02 2 .28 .02 .02 .01 For your convenience, I have calculated the following: ?(?)=6.24,?(?2)=40.34,?(?)=0.89,?(?2)=1.55E(X)=6.24,E(X2)=40.34,E(Y)=0.89,E(Y2)=1.55. A.    Determine the marginal distribution of X. B.    Calculate E(XY). C.    Calculate the correlation between X...
3. Assume that Rahim has $168 to spend on books and movies each month and that...
3. Assume that Rahim has $168 to spend on books and movies each month and that both goods must be purchased whole (no fractional units). Movies (X) cost $12 and books (Y) cost $30 each on an average. Rahim’s preferences for movies and books are summarized by the following information: No. per Month Movies No. per Month Books TU MU MU/Px TU MU MU/Py 1 56 1 150 2 94 2 270 3 124 3 360 4 148 4 410...
We know that the population distribution for the number of hours that Americans sleep per night...
We know that the population distribution for the number of hours that Americans sleep per night is normally distributed with a mean of 6.7 and a standard deviation of 1.4. a) What is the probability that a single, randomly draw observation (i.e., American) from this distribution sleeps less than 5 hours per night? b) What is the probability that a randomly drawn observation falls between 6 and 8 hours per night? c) What is the probability that a randomly draw...
The population mean number of hours a group watched TV was 10.5 with a standard deviation...
The population mean number of hours a group watched TV was 10.5 with a standard deviation of 3.6. A random sample of 36 individuals from the group was selected and the number of hours each watched TV was obtained. Consider the statistic the sample mean number of hours the studied group of 36 watched TV. Question 6 options: 12345678 Consider the statistic the sample mean number of hours the studied group of 36 watched TV. What is the mean of...
ADVERTISEMENT
ADVERTISEMENT
ADVERTISEMENT