Question

A researcher wants to see if a kelp extract helps prevent frost damage on tomato plants. One hundred tomato plants in individual containers are randomly assigned to two different groups. Plants in both groups are treated identically, except that the plants in group 1 are sprayed weekly with a kelp extract, while the plants in group 2 are not. After the first frost, 12 of the 50 plants in group 1 exhibited damage and 18 of the 50 plants in group 2 showed damage. Let p1 be the actual proportion of all tomato plants of this variety that would experience damage under the kelp treatment, and let p2 be the actual proportion of all tomato plants of this variety that would experience damage under the no-kelp treatment.

Required:
Construct and interpret a 95% confidence interval for the true difference in the proportion of tomato plants like these that would experience damage after receiving the kelp treatment and no-kelp treatment.

Answer 1

The 95% confidence interval for the true difference in the proportion of tomato plants like these that would experience damage after receiving the kelp treatment and no kelp treatment is -0.2690 <
`\hat{p}_1-\hat{p}_2` < 0.02901

Explanation:

The given data are;

The number of the group 1 plants that exhibited damage = 12

The number of plants in group 1, n₁ = 50

The number of the group 2 plants that exhibited damage = 18

The number of plants in group 2, n₂ = 50

The proportion of group 1 plants that exhibited damage,
`\hat p_1` = 12/50 = 0.24

The proportion of group 2 plants that exhibited damage,
`\hat p_2` = 18/50 = 0.36

The z-value at 95% = 1.64

The confidence interval is given by the following formula;

`C.I. = \hat{p}_1-\hat{p}_2\pm z^(*)\sqrt{\frac{\hat{p}_1\left (1-\hat{p}_1 \right )}{n_(1)}+\frac{\hat{p}_2\left (1-\hat{p}_2 \right )}{n_(2)}}`

Plugging in the values, we get;

`C.I. = 0.24-0.36\pm 1.64 * \sqrt{(0.24\left (1-0.24 \right ))/(50)+(0.36\left (1-0.36 \right ))/(50)}`

Therefore, C.I. = -0.2690 <
`\hat{p}_1-\hat{p}_2` < 0.02901