The distribution of heights of a certain breed of terrier has a mean of 72 cm. a

Question

The distribution of heights of a certain breed of terrier has a mean of 72 cm. and a standard deviation of 10 cm, whereas the distribution of heights of a certain breed of poodle has a mean of 28 cm with a standard deviation of 5 cm. Assuming that the sample means can be measured to any degree of accuracy, find the probability that the sample mean for a random sample of heights of 64 terriers exceeds the sample mean for a random sample of heights of 100 poodles by at most 44.2 centimeters. A manufacturing firm claims that the batteries used in their electronic games will last an average of 30 hrs. To maintain this average, 16 batteries are tested each month. If the computed t-value falls between -to.o25 and t_0 025, the firm is satisfied with its claim. What conclusion should the firm draw from a sample that has a mean of x = 27.5 hrs. and a SD of s = 5 hrs? Assume the distribution of battery lives to be approximately normal. Given a normal random variable X with mu = 20 and simga^2 = 9, and a random sample n taken from the distribution, what sample size n is necessary in order that P(19.9

Explanation / Answer

1. As per the problem, we have two separate populations.
    For population 1 (Terriers), µ1 = 72 and 1 = 10
    For population 2 (Poodles), µ2 = 28 and 2 = 5

We have to find the sampling distribution of the difference between means of the two populations.

The mean of the sampling distribution of the difference between the means = µ1-2
    = µ1 - µ2
   = 72 – 28
   = 44

The standard error of the sampling distribution of the difference between means is given by 1-2
    = [ 12 / n1 + 22 / n2 ]1/2
    = [ 102 / 64 + 52 / 100 ]1/2
    = [ 100 / 64 + 25 / 100 ] 1/2
    = 1.3463

The sampling distribution of the difference between means will be normal with Mean 44 and S.D. 1.3463.

The z-value (standard normal value) for 44.2 = (44.2 - 44) / 1.3463 = 0.1486
Therefore, the required probability is the cumulative probability for a z-value of 0.1486 0.15
From the standard normal table, the required probability is 0.55962.

2. Here, we are drawing a small sample (n = 16) from a normal population. Therefore, the sampling distribution of the sample mean will follow a t-distribution.

The test statistic can be calculated as:
t = ( 27.5 – 30 ) / ( 5 / 16 )
= -2.5 / 1.25
= -2

Now, from the t-distribution table for n-1 = 15 degrees of freedom, t0.025 = 2.131.
Since the test statistic (i.e. -2) lies between -2.131 and 2.131, the firm can be satisfied with its claim.

3. A standard normal distribution contains approximately 95% of the cumulative probability within two standard deviations of the mean.

If a random sample is pulled from a normal population, the sampling distribution of the sample mean has a mean equal to that of the population. Therefore, for P(19.9 Mean(x) 20.1) = 0.95,
(20 -19.9) = 0.1 should be equal to 2 standard deviations (i.e. standard errors) of the sampling distribution of x.

Standard Error of the sample mean = / n = 3 / n

Therefore, 3 / n = 0.1 /2
   or, n = 3 / 0.05 = 60
  or, n = 3600

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