A Kansas University research station wishes to investigate if a new variety of wheat is more resistant to a disease than an old variety. It is known that this disease strikes approximately 15% of all plants of the old variety. A field experiment was conducted, and of 120 new plants, 12 became infected.
The calculated value of the test statistic is:
1.83
1.10
-1.53
1.53
1.83
To calculate the test statistic, we can use the formula for a one-sample proportion test:
test statistic = (p̂ - p) / √(p * (1 - p) / n)
where:
p̂ is the proportion of infected plants in the new variety (12/120 = 0.10)
p is the proportion of infected plants in the old variety (15% or 0.15)
n is the sample size (120)
Substituting the values into the formula:
test statistic = (0.10 - 0.15) / √(0.15 * (1 - 0.15) / 120)
Simplifying the equation:
test statistic = -0.05 / √(0.15 * 0.85 / 120)
Calculating:
test statistic = -0.05 / √(0.1275 / 120)
test statistic = -0.05 / √(0.0010625)
test statistic = -0.05 / 0.0326046
Finally, calculating the test statistic:
test statistic ≈ -1.532 (rounded to three decimal places)
Therefore, the calculated value of the test statistic is -1.53 when rounded to two decimal places.
To calculate the test statistic, we need to use a hypothesis test called the two-proportion z-test. This test is used to compare the proportions of two independent groups to determine if they are significantly different.
In this case, we are comparing the proportion of infected plants in the new variety to the proportion of infected plants in the old variety. Let's denote the proportion of infected plants in the new variety as p1 and the proportion of infected plants in the old variety as p2.
The null hypothesis, denoted as H0, assumes that there is no difference in disease resistance between the two varieties. So, H0: p1 = p2.
The alternative hypothesis, denoted as Ha, assumes that there is a difference in disease resistance between the two varieties. So, Ha: p1 ≠ p2.
To calculate the test statistic, we use the formula:
z = (p1 - p2) / √(p0 * (1-p0) * (1/n1 + 1/n2))
where p0 is the pooled proportion, calculated as (x1 + x2) / (n1 + n2), and x1 and x2 are the number of infected plants in each variety and n1 and n2 are the corresponding total number of plants.
In this case, x1 = 12, x2 = 15% of 120 = 18, n1 = 120, and n2 = 120.
So, the pooled proportion can be calculated as p0 = (12 + 18) / (120 + 120) = 30 / 240 = 0.125.
Plugging the values into the formula, we have:
z = (0.125 - 0.15) / √(0.125 * (1 - 0.125) * (1/120 + 1/120))
z = -0.025 / √(0.125 * 0.875 * (1/120 + 1/120))
z = -0.025 / √(0.125 * 0.875 * 0.01667)
z = -0.025 / √0.0014583333
z = -0.025 / 0.0381671918
z ≈ -0.6552
So, the calculated value of the test statistic is -0.6552.
Therefore, the correct answer is: -0.6552.
You can try a proportional one-sample z-test for this one since this problem is using proportions.
Using a formula for a proportional one-sample z-test with your data included, we have:
z = .1 - .15 -->test value (12/120) is .1) minus population value (.15) divided by
√[(.15)(.85)/120] --> .85 represents 1-.15 and 120 is sample size.
Finish the calculation.