Answer to Question #162726 in Statistics and Probability for Sunny

Question

Data from department of agriculture in a country show that average size of the farms has increased since 1975. In 1975 the mean size of the farm was 471 acres. an agribusiness researcher believes that average size of the farm has increased. To test this a random sample of 23 farms across the country was taken. The sample mean was found to be 498.78 acres with a standard deviation of 46.94. using 5% level of significance, test whether agribusiness researcher's belief is true or not?

Answer 1

We have that

$\mu=471$

$n=23$

$\bar x=498.78$

$s=46.94$

$\alpha=0.05$

$H_0:\mu = 471$

$H_a:\mu >471$

The hypothesis test is right-tailed.

Since the population standard deviation is unknown we use the t-test.

The critical value for 5% significance level and 22 df is 1.717

(degrees of freedom df = n – 1 = 23 – 1 = 22)

The critical region is t > 1.717

Test statistic:

$$t=\frac{\bar x - \mu}{\frac{s}{\sqrt n}}=\frac{498.78- 471}{\frac{46.94}{\sqrt {23}}}=2.838$$

Since 2.838 > 1.717 thus t falls in the rejection region we reject the null hypothesis.

At the 5% significance level the data do provide sufficient evidence to support the agribusiness researcher's belief. We are 95% confident to conclude that the average size of the farm has increased.