Question

The following data represent petal lengths (in cm) for independent random samples of two species of Iris.

Petal length (in cm) of Iris virginica: x₁; n₁ = 35

5.1	5.9	6.1	6.1	5.1	5.5	5.3	5.5	6.9	5.0	4.9	6.0	4.8	6.1	5.6	5.1
5.6	4.8	5.4	5.1	5.1	5.9	5.2	5.7	5.4	4.5	6.4	5.3	5.5	6.7	5.7	4.9
4.8	5.9	5.1

Petal length (in cm) of Iris setosa: x₂; n₂ = 38

1.5	1.9	1.4	1.5	1.5	1.6	1.4	1.1	1.2	1.4	1.7	1.0	1.7	1.9	1.6	1.4
1.5	1.4	1.2	1.3	1.5	1.3	1.6	1.9	1.4	1.6	1.5	1.4	1.6	1.2	1.9	1.5
1.6	1.4	1.3	1.7	1.5	1.7

(a) Use a calculator with mean and standard deviation keys to calculate x₁, s₁, x₂, and s₂. (Round your answers to two decimal places.)

x1 =
s1 =
x2 =
s2 =

(b) Let μ₁ be the population mean for x₁ and let μ₂ be the population mean for x₂. Find a 99% confidence interval for μ₁ − μ₂. (Round your answers to two decimal places.)

lower limit
upper limit

(c) Explain what the confidence interval means in the context of this problem. Does the interval consist of numbers that are all positive? all negative? of different signs? At the 99% level of confidence, is the population mean petal length of Iris virginica longer than that of Iris setosa?

Because the interval contains only positive numbers, we can say that the mean petal length of Iris virginica is longer.Because the interval contains only negative numbers, we can say that the mean petal length of Iris virginica is shorter.     Because the interval contains both positive and negative numbers, we cannot say that the mean petal length of Iris virginica is longer.

(d) Which distribution did you use? Why?

The standard normal distribution was used because σ₁ and σ₂ are unknown.The Student's t-distribution was used because σ₁ and σ₂ are known.     The standard normal distribution was used because σ₁ and σ₂ are known.The Student's t-distribution was used because σ₁ and σ₂ are unknown.

Do you need information about the petal length distributions? Explain.

Both samples are large, so information about the distributions is needed.Both samples are large, so information about the distributions is not needed.     Both samples are small, so information about the distributions is needed.Both samples are small, so information about the distributions is not needed.

Answer 1

X1
5.1 5.9 6.1 6.1 5.1 5.5 5.3 5.5 6.9 5.0 4.9 6.0 4.8 6.1 5.6 5.1 5.6 4.8 5.4 5.1 5.1 5.9 5.2 5.7 5.4 4.5 6.4 5.3 5.5 6.7 5.7 4.9 4.8 5.9 5.1	0.1521 0.1681 0.3721 0.3721 0.1521 1E-04 0.0361 1E-04 1.9881 0.2401 0.3481 0.2601 0.4761 0.3721 0.0121 0.1521 0.0121 0.4761 0.0081 0.1521 0.1521 0.1681 0.0841 0.0441 0.0081 0.9801 0.8281 0.0361 1E-04 1.4641 0.0441 0.3481 0.4761 0.1681 0.1521
= 192	= 10.7035

a) Mean of X1 =

Sample standard deviation of X1 =

X2
1.5 1.9 1.4 1.5 1.5 1.6 1.4 1.1 1.2 1.4 1.7 1.0 1.7 1.9 1.6 1.4 1.5 1.4 1.2 1.3 1.5 1.3 1.6 1.9 1.4 1.6 1.5 1.4 1.6 1.2 1.9 1.5 1.6 1.4 1.3 1.7 1.5 1.7	0.0001 0.1681 0.0081 0.0001 0.0001 0.0121 0.0081 0.1521 0.0841 0.0081 0.0441 0.2401 0.0441 0.1681 0.0121 0.0081 0.0001 0.0081 0.0841 0.0361 0.0001 0.0361 0.0121 0.1681 0.0081 0.0121 0.0001 0.0081 0.0121 0.0841 0.1681 0.0001 0.0121 0.0081 0.0361 0.0441 0.0001 0.0441
= 56.8	= 1.7398

Mean of X2 =

sample standard deviation of X2 =

Pooled standard deviation is given by -

b) 99% confidence interval for is given by -

= [3.77, 4.23]

Lower Limit = 3.77

Upper Limit = 4.23

c) Because the interval contains only positive numbers, we can say that the mean petal length of the Iris virginica is longer.

d) As the population standard deviation of the population is unknown, t distribution was used.

So,the answer will be - The Student's t distribution was used because are unknown.

Since, both the samples are Large, so, the information about the distribution is not needed.