Experience raising New Jersey Red chickens revealed the mean weight of the chickens at five months is 4.35 pounds. The weights follow the normal distribution. In an effort to increase their weight, a special additive is added to the chicken feed. The subsequent weights of a sample of five-month-old chickens were (in pounds): (In all your following computations, use three decimal places.)

4.41 4.37 4.33 4.35 4.30 4.39 4.36 4.38 4.40 4.39

a. What is the sample mean of the weights of the chickens after being fed the new additive?

b. Would you use the Z-distribution or the Student-t distribution to develop a confidence interval for these data? Explain your choice.

c. Having made your selection of the appropriate probability distribution in part (b), what is the value of Z (or t) for a 95% confidence interval to test whether the additive has increased the weight of the chickens?

d. What is the 95% confidence interval for the population mean weight of the chickens after they were fed the new additive?

e. Based on your 95% confidence interval, discuss whether you can conclude that the mean weight of the chickens has statistically increased.

f. Does it make any difference to your conclusion in part (e) whether you used the normal distribution or the t-distribution? Explain.

g. Does it make any difference to your conclusion in part (e) whether you used a 90% confidence interval instead of a 95% confidence interval? Explain

Solution Summary

The solution examines statistics for New Jersey Red Chickens.

... In this problem, we find the mean and standard deviation of the sample mean, then standardize (convert to a Z distribution) by subtracting the mean and then ...

... a z-distribution table, we can see that the area between z = -1.449 and z = 0.942 is 0.4265 + 0.3264 = 0.7529. This means that 75.29% of the sample means will ...

... For example, 1 is converted to az value of 1.00. ... Note that the center of the z distribution is zero, indicating no deviation from the mean, . ...

... What is the probability that the sample mean height is greater than inches? Cannot be answered. We cannot use z distribution because the sample size is less ...

... are: Sample mean = 25.4 = Population standard deviation = 3.1 Sample size = 51 = The 95th percentile of the standard normal (Z) distribution = 1.645 When we ...

... of the z-distribution? Always has a mean of 0 and standard deviation of 1. 2. Define the central limit theorem: It is how a distribution of sample means is a ...

... From the z-distribution table, the critical value of the test statistic for is given as. ... This indicates that our sample mean of 65 is not an extreme or unusual ...

... that the sample mean will be 25000 or more? For all of these problems, we're going to assume that the distributions are normal and use the z-distribution. ...

... is known, use normal distribution Thus we use z distribution Z at ... Reject) Null Hypothesis 3) Calculation of sample statistics Hypothesized Mean=μ ...

... 2. It is, like the z distribution, bell-shaped and symmetrical. ... All t distri- butions have a mean of 0 ... standard deviations differ according to the sample size, n ...