Question

In: Statistics and Probability

(a). A simple random sample of 37 days was selected. For these 37 days, Parsnip was...

(a). A simple random sample of 37 days was selected. For these 37 days, Parsnip was fed seeds on 22 days and Parsnip was fed pellets on the other 15 days. The goal is to calculate a 99% confidence interval for the proportion of all days in which Parsnip was fed seeds, and to do so there are two assumptions. The first is that there is a simple random sample, which is satisfied. What is the second assumption, and specific to the information in this problem, is it satisfied? Explain why or why not.

(b). If appropriate, use the data in part (a) to calculate and interpret a 99% confidence interval for the proportion of all days in his life that Parsnip has been fed seeds.

Solutions

Expert Solution

solution:

the given sample size = n=37

number of days on which parship was fed feed = X = 22

so proportion of sample = = 22/37 = 0.595

condition:

1: it was given that the sample is a random sample so random condition is satisfied

2: independent condition : the sample value is assumed to be independent of each other.

3: 10% conditin: the sample size should not be more than 10% of the total population so this condition also satisfied because there are many parship in the population is asssumed.

b) if our all the condition are satisfied then it is appropriate to find the confidence interval so the confidence interval are calculated as follows:

significance level = = 0-0.99 = 0.01

critical value of z =

margin of error =

confidence interval = ( - E) < p < ( + E)

upper limit = 0.595 + 0.208 = 0.803

lower limit = 0.595 - 0.208 = 0.387

interpretation:

with 99 % confident that the true proportion of parsnip was fed seed is between 0.387 and 0.803 .


Related Solutions

(a). A simple random sample of 37 days was selected. For these 37 days, Parsnip was...
(a). A simple random sample of 37 days was selected. For these 37 days, Parsnip was fed seeds on 22 days and Parsnip was fed pellets on the other 15 days. The goal is to calculate a 99% confidence interval for the proportion of all days in which Parsnip was fed seeds, and to do so there are two assumptions. The first is that there is a simple random sample, which is satisfied. What is the second assumption, and specific...
(a). A simple random sample of 37 days was selected. For these 37 days, Parsnip was...
(a). A simple random sample of 37 days was selected. For these 37 days, Parsnip was fed seeds on 20 days and Parsnip was fed pellets on the other 17 days. The goal is to calculate a 95% confidence interval for the proportion of all days in which Parsnip was fed seeds, and to do so there are two assumptions. The first is that there is a simple random sample, which is satisfied. What is the second assumption, and specific...
1. A simple random sample of 22 days in which Parsnip ate seeds was selected, and...
1. A simple random sample of 22 days in which Parsnip ate seeds was selected, and the mean amount of time it took him to eat the seeds was 15.3 seconds with a standard deviation of 4.01 seconds. An independent simple random sample of 15 days in which Parsnip ate pellets was selected, and the mean amount of time it took him to eat the pellets was 14.2 seconds with a standard deviation of 3.89 seconds. If appropriate, use this...
A simple random sample of 300 colleges and universities was selected and for each it was...
A simple random sample of 300 colleges and universities was selected and for each it was determined whether they planned to move summer 2020 classes online or not. Of the 300 colleges and universities in the sample, 90 responded that they will move summer 2020 classes online , and 210 responded that they will not move summer 2020 classes online. . If appropriate, use this information to calculate and interpret a 99% confidence interval for the proportion of all colleges...
A simple random sample of germination times (in days) for 7 seeds of a new strain...
A simple random sample of germination times (in days) for 7 seeds of a new strain of snap beans has a sample mean of 15.57 and standard deviation of 3.21. (a) Determine a 99% confidence interval for the true mean germination time for this strain. (b) Choose a correct interpretation to your result in part (a): i. We can be 99% sure the sample average of 7 germination times is in the interval calculated in part (a). ii. We can...
A simple random sample of 101 cities and counties in the United States was selected and...
A simple random sample of 101 cities and counties in the United States was selected and the number of positive COVID-19 cases per 1000 citizens was recorded for each. The mean number of positive cases for this sample of 101 counties and cities was 253, with a standard deviation of 11.4. Due to a couple of cities with extremely high numbers of cases the distribution is skewed heavily to the right. If appropriate, use this information to calculate and interpret...
A simple random sample of 81 is selected from a population with a standard deviation of...
A simple random sample of 81 is selected from a population with a standard deviation of 17. The degree of confidence is 90%. What is the margin of error for the mean?
A simple random sample of size n=37 is obtained from a population with μ=65 and σ=15....
A simple random sample of size n=37 is obtained from a population with μ=65 and σ=15. ​(a) What must be true regarding the distribution of the population in order to use the normal model to compute probabilities involving the sample​ mean? Assuming that this condition is​ true, describe the sampling distribution of overbarx. ​(b) Assuming the normal model can be​ used, determine ​P(overbar x < 69.4). ​(c) Assuming the normal model can be​ used, determine ​P(overbar x ≥ 66.7​).
Suppose a simple random sample of 90 students from this district is selected. What is the...
Suppose a simple random sample of 90 students from this district is selected. What is the probability that at least half of them have brown eyes? (Use technology. Round your answer to three decimal places.) Brown eyes is 40% I got 0.026 as my answer but the program is saying it is wrong, can someone figure out the correct answer please
A simple random sample of 80 colleges and universities is selected, and 16 indicated will move...
A simple random sample of 80 colleges and universities is selected, and 16 indicated will move their summer 2020 classes online while the other 64 indicated they did not plan to move their summer 2020 classes online . If appropriate, use this information to test the hypotheses stated in question 10 at the a = .10 level of significance. Please type your answer in the box below. Hypothesis: H0:   π = 0.14 versus HA: π > 0.14
ADVERTISEMENT
ADVERTISEMENT
ADVERTISEMENT