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Karo-lina-s [1.5K]
3 years ago
7

The average size of 8 farms in Indiana County, Pennsylvania, is 191 acres with standard deviation of 38 acres. The average size

of 10 farms in Greene County, Pennsylvania, is 199 acres with standard deviation of 12 acres. Can it be concluded that average size of the farms in Indiana County and the average size of the farms in Greene County are different
Mathematics
1 answer:
Grace [21]3 years ago
4 0

Answer:

Step-by-step explanation:

This is a test of 2 independent groups. The population standard deviations are not known. it is a two-tailed test. Let μ1 be average size of farms in Indiana County, Pennsylvania and μ2 be the average size of farms in Greene County, Pennsylvania.

The random variable is μ1 - μ2 = difference in the average size of the farms in Indiana County and the average size of the farms in Greene County.

We would set up the hypothesis.

The null hypothesis is

H0 : μ1 = μ2 H0 : μ1 - μ2 = 0

The alternative hypothesis is

H1 : μ1 ≠ μ2 H1 : μ1 - μ2 ≠ 0

Since sample standard deviation is known, we would determine the test statistic by using the t test. The formula is

(μ1 - μ2)/√(s1²/n1 + s2²/n2)

From the information given,

μ1 = 191

μ2 = 199

s1 = 38

s2 = 12

n1 = 8

n2 = 10

t = (191 - 199)/√(38²/8 + 12²/10)

t = - 0.57

The formula for determining the degree of freedom is

df = [s1²/n1 + s2²/n2]²/(1/n1 - 1)(s1²/n1)² + (1/n2 - 1)(s2²/n2)²

df = [38²/8 + 12²/10]²/[(1/8 - 1)(38²/8)² + (1/10 - 1)(12²/10)²] = 37986.01/4677.36142857143

df = 8

We would determine the probability value from the t test calculator. It becomes

p value = 0.58

Let us assume a significance level of 0.05

Since alpha, 0.05 < than the p value, 0.58, then we would accept the null hypothesis. Therefore, at a significance level of 5%, there is not sufficient evidence to conclude that the average size of the farms in Indiana County and the average size of the farms in Greene County are different.

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