Question

The distribution of pregnancy lengths of a certain mammal is known to be right skewed with a population mean of 170 days and a population standard deviation of 100 days.

a. Suppose four random samples with different sample size are drawn from this distribution and the sample mean for each is calculated. The same means were 145,169.8,183.3,175. Match the appropriate sample size below with each sample mean, using the Law of Large Numbers.

Sample Size 2..... Sample Mean =
Sample Size 20.... Sample Mean =
Sample Size 50.... Sample Mean =
Sample Size 200... Sample Mean =


b. Suppose we repeatedly take samples of size 100 from the population distribution, calculate a sample mean each time, and plot those sample means in a histogram. The histogram we created would be an example of a  --- sampling distribution variable population distribution . According to the central limit theorem, the histogram would have a shape that is approximately  --- right skewed normal left skewed , with mean  (give a number) and standard deviation  (give a number). The standard deviation of the statistic under repeated sampling is called the  --- standard error absolute error deviation absolute deviation . The middle 95% of the histogram we created lies between  and  (give numbers for both blanks with the smaller number listed first).


c. Suppose we draw one sample of size 100 from the population distribution of animal pregnancy lengths.
(Round to 2 decimal places)
I. Calculate a z-score for a sample mean of 175.2.
II. Calculate a z-score for a sample mean of 152.
III. Is it more likely that we would observe a sample mean of 175.2 or 152?  --- equally likely 152 175.2
Justify your answer to III in one sentence.

Answer 1

(a)

Suppose, random variable X denotes pregnancy length of certain mammal.

Expected mean is 170 days. It is achieved with increase of sample size.

So, most deviation from 170 of our observed mean values is expected to occur in case of lowest sample size and it decreases with increase of sample size.

Hence,

• Sample Size = 2. Sample Mean = 145
• Sample Size = 20. Sample Mean = 183
• Sample Size = 50. Sample Mean = 175
• Sample Size = 200. Sample Mean = 169

(b)

The histogram we created would be an example of a sampling distribution.

According to the central limit theorem, the histogram would have a shape that is approximately normal, with mean 170 and standard deviation 10.

The standard deviation of the statistic under repeated sampling is called the standard error.

We know,

The middle 95% of the histogram we created lies between 150.4004 and 189.5996.

(c)

By the central limit theorem (CLT), .

To get corresponding z-score we have to perform the transformation,

(i)

For ,

z = (175.2-170) / 10 = 0.52

(ii)

For ,

z = (152-170) / 10 = -1.8

(iii)

Sample mean with value nearer to 170 is more likely.

Hence, between these two 175.2 is more likely.