Question

In: Statistics and Probability

A sample of 8 mountain goats has sample mean weight 85.6 lbs and sample standard deviation...

A sample of 8 mountain goats has sample mean weight 85.6 lbs and sample standard deviation s=20.3 lbs. Find a 95% confidence interval for the mean weight of the population mu.

Solutions

Expert Solution

Solution :

Given that,

Point estimate = sample mean = = 85.6

sample standard deviation = s = 20.3

sample size = n = 8

Degrees of freedom = df = n - 1 = 8-1 =7

At 95% confidence level the t is ,

= 1 - 95% = 1 - 0.95 = 0.05

/ 2 = 0.05 / 2 = 0.025

t /2,df = t0.025,7= 2.365

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

= 2.365 * (20.3 / 8)

Margin of error = E = 17

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

- E < < + E

85.6 - 17 < < 85.6 + 17

68.6 < < 102.6

(68.6,102.6)


Related Solutions

In a sample of 51 babies, the mean weight is 21 pounds and the standard deviation...
In a sample of 51 babies, the mean weight is 21 pounds and the standard deviation is 4 pounds. Calculate a 95% confidence interval for the true mean weight of babies. Suppose we are interested in testing if the true mean weight of babies is 19.4 vs the alternative that it is not 19.4 with an alpha level of .05. Would this test be significant? Explain your answer. Perform the t test and use a t-table to get the p-value
A sample of 16 toy dolls had a mean weight of 71.5 and a standard deviation...
A sample of 16 toy dolls had a mean weight of 71.5 and a standard deviation of 12 pounds, respectively. Assuming normality, construct 95% confidence interval for the population mean weight, μ.
In a sample of 36 suitcases , the avg is 45 lbs. Standard deviation is 8.2....
In a sample of 36 suitcases , the avg is 45 lbs. Standard deviation is 8.2. Construct at 90% confidence interval for the mean of the weight of all suticases
Suppose cattle in a large heard have a mean weight of 1058 lbs and a standard...
Suppose cattle in a large heard have a mean weight of 1058 lbs and a standard deviation of 90. What is teh probablity that the mean weight of a sample of steers would be 12 or more pounds less than the population mean if 51 steers are samples at random from the herd?
A population has a mean of 180 and a standard deviation of 36. A sample of...
A population has a mean of 180 and a standard deviation of 36. A sample of 84 observations will be taken. The probability that the sample mean will be between 181 and 185 is
A population has a mean of 180 and a standard deviation of 24. A sample of...
A population has a mean of 180 and a standard deviation of 24. A sample of 64 observations will be taken. The probability that the sample mean will be between 183 and 186 is
A population has a mean of 37.6 and a standard deviation of 14.8. A sample of...
A population has a mean of 37.6 and a standard deviation of 14.8. A sample of 75 will be taken. Find the probability that the sample mean will be greater than 42.0. a) Calculate the z score. (Round your answer to 2 decimals.) b) Find the probability that the sample mean will be greater than 42.0. (Round your answer to 4 decimals.)
A population has a mean of 245.3 and a standard deviation of 12.6. A sample of...
A population has a mean of 245.3 and a standard deviation of 12.6. A sample of 200 will be taken. Find the probability that the sample mean will be less than 248.4. a) Calculate the z score. (Round your answer to 2 decimals.) b) Find the probability that the sample mean will be less than 248.4. (Round your answer to 4 decimals.)
A population has a mean of 300 and a standard deviation of 18. A sample of...
A population has a mean of 300 and a standard deviation of 18. A sample of 144 observations will be taken. The probability that the sample mean will be less than 303 is 0.4332 0.9772 0.9544 0.0668 None of the above
A population has a mean of 180 and a standard deviation of 24. A sample of...
A population has a mean of 180 and a standard deviation of 24. A sample of 64 observations will be taken. The probability that the sample mean will be within 3 of the population mean is:
ADVERTISEMENT
ADVERTISEMENT
ADVERTISEMENT