Question

In: Statistics and Probability

A researcher selects a sample of 100 participants from a population with a mean of 38...

A researcher selects a sample of 100 participants from a population with a mean of 38 and a standard deviation of 20. About 68% of the sample means in this sampling distribution should be between ______ and ______. Show your work

Solutions

Expert Solution

Solution :

Given that,

= 38

= 20

Using Empirical rule,

= 38 and

= / n = 20 / 100 = 20 / 10 = 2

Using Empirical rule,

P(  - 1 < <   + 1 ) = 68%

P(38 - 1 * 2 < < 38 + 1 * 2) = 68%

P(36 < < 40) = 68%

About 68% of the sample means in this sampling distribution should be between 36 and 40 .


Related Solutions

A researcher selects a sample of n=25 individuals from a population with a mean of μ=60...
A researcher selects a sample of n=25 individuals from a population with a mean of μ=60 and standard deviation of σ=10 and administers a treatment. The researcher predicts that the treatment will increase scores. Following the treatment, the average scores for this sample is M=65. 1. Using symbols, state the hypothesis for a one-tailed test. 2. Calculate the z-score and place this on a standardized normal distribution. 3. With a one-tail α=0.05, state the conclusion of these findings. 4. Calculate...
. A researcher selects a random sample of 10 persons from a population of truck drivers...
. A researcher selects a random sample of 10 persons from a population of truck drivers and gives them a driver’s aptitude test. Their scores are 22,3,14,8,11,5,18,13,12, and 12.      (a) Find the estimated standard error of the mean.      (b) Find the 95% confidence interval for the population mean.
Question 27 options: A researcher selects a sample of n = 25 from a normal population...
Question 27 options: A researcher selects a sample of n = 25 from a normal population with µ = 80 and σ = 20. If the treatment is expected to increase scores by 6 points, what is the power of a two-tailed hypothesis test using α = .05? Enter the result for each step below: Step 1: Enter the standard error, σM (enter a number with 5 decimal places using only the keys "0-9" and "."): Step 2: Enter the...
A researcher selects 100 subjects at random from a population, observes 50 successes, and calculates three...
A researcher selects 100 subjects at random from a population, observes 50 successes, and calculates three confidence intervals. The confidence levels are 90%, 95%, and 99%, and the intervals are (0.402, 0.598), (0.371, 0.629), and (0.418, 0.582). Match each interval with its confidence level.
A sample of 38 observations is selected from a normal population. The sample mean is 47,...
A sample of 38 observations is selected from a normal population. The sample mean is 47, and the population standard deviation is 7. Conduct the following test of hypothesis using the 0.05 significance level. H0: μ =48 H1: μ ≠ 48 What is the decision rule? Reject H0 if z < -1.960 or z > 1.960 A.) What is the value of the test statistic? (Negative amount should be indicated by a minus sign. Round your answer to 2 decimal...
2. A sample of 100 is selected from a population with mean µ = 100 and...
2. A sample of 100 is selected from a population with mean µ = 100 and σ = 20.0. (a) Describe the sampling distribution of the sample mean X¯. (b) Find P(X >¯ 105). (c) Find P(100 < X <¯ 105). (d) Find P(98 < X <¯ 104). 3. According to the U.S. Department of Energy, the average price of unleaded regular gasoline sold at service stations throughout the nation is $1.23 per gallon with a standard deviation of $0.16....
Sixty items are randomly selected from a population of 660 items. The sample mean is 38,...
Sixty items are randomly selected from a population of 660 items. The sample mean is 38, and the sample standard deviation 3. Develop a 98% confidence interval for the population mean. (Round the final answers to 2 decimal places.)                  The confidence interval is between       and     .
A random sample of 100 observations from a quantitative population produced a sample mean of 29.8...
A random sample of 100 observations from a quantitative population produced a sample mean of 29.8 and a sample standard deviation of 7.2. Use the p-value approach to determine whether the population mean is different from 31. Explain your conclusions. (Use α = 0.05.) State the null and alternative hypotheses. (Choose Correct Letter) (a) H0: μ = 31 versus Ha: μ < 31 (b) H0: μ ≠ 31 versus Ha: μ = 31     (c) H0: μ < 31 versus Ha:...
A random sample of n =100 is selected from a normal population with mean μ =...
A random sample of n =100 is selected from a normal population with mean μ = 24 and standard deviation σ = 1.25. Find the probability that  is less than 24.3
A random sample of n = 100 observations is drawn from a population with mean 10...
A random sample of n = 100 observations is drawn from a population with mean 10 and variance 400. (a) Describe the shape of the sampling distribution of ¯x. Does your answer depend on the sample size? ( b) Give the mean and standard deviation of the sampling distribution of ¯x. (c) Find the probability that ¯x is less than 8. (d) Find the probability that ¯x is greater than 9.5. (e) Find the probability that ¯x is between 8...
ADVERTISEMENT
ADVERTISEMENT
ADVERTISEMENT