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Someone is researching different fertilizers and their effect on the growth of plants. The researcher tests 8 pairs of plants, all of the same type. They give one plant in each pair fertilizer A and the other plant fertilizer B. They hypothesize that there will be a difference in the growth of the plants but are unsure which fertilizer will be better. The researcher finds that when they compare plant growth, [tex]\bar{D} = 8.04[/tex] and the variance is [tex]s_d^2 = 79.22[/tex]. Plants given fertilizer A grew to 13 cm, while plants given fertilizer B grew to 8 cm.

1. Which type of t-test is this?
2. What are the t-obtained (t-obt) and t-critical (t-crit) values?
3. What is the conclusion?

Answer :

This is a paired t-test as the researcher is comparing the growth of plants within each pair, where one plant is given fertilizer A and the other plant is given fertilizer B.

To calculate the t-obtained, we use the formula:

t-obt = (Dbar - μD) / (sD / √n)

where Dbar is the mean difference in growth (8.04), μD is the hypothesized mean difference (0 since we assume no difference), sD is the standard deviation of the differences (square root of s^2d), and n is the number of pairs (8).

To determine the t-critical value, we need to specify the significance level (α). Let's assume α = 0.05 and degrees of freedom (df) = n - 1 = 7. Using a t-table or statistical software, we can find the t-critical value.

The conclusion depends on comparing the t-obtained to the t-critical value. If t-obtained > t-critical, we reject the null hypothesis and conclude that there is a significant difference in plant growth between the two fertilizers. If t-obtained ≤ t-critical, we fail to reject the null hypothesis and conclude that there is not enough evidence to suggest a significant difference in plant growth.

To know more about t-test click here:

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