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Professor Sprout is researching the impact of using different fertilizers on plant growth. She wants to control for the type of plant, so she tests 8 pairs of plants of the same type. She gives one plant in each pair fertilizer A and the other plant fertilizer B. She thinks there will be a difference in the growth of the plants, but she is not sure which will be better. She finds that when she compares plant growth, [tex]D_{\text{bar}} = 8.04[/tex] and the variance is [tex]S^2_D = 79.22[/tex]. Plants given fertilizer A grew to 13 cm, while plants given fertilizer B grew to 8 cm. Use an alpha of 0.05.

Run the appropriate test and report the following:

1. Which type of t-test is this?
2. What are your degrees of freedom?
3. What is your t-critical value?
4. What is your t-obtained value?
5. What is your conclusion in APA format with statistical evidence? Do you reject or fail to reject the null hypothesis?

Answer :

The Professor Sprout's research showed that the type of fertilizer used had a significant impact on the growth of plants. Fertilizer A resulted in significantly higher growth compared to fertilizer B.

This is a paired samples t-test.

Degrees of freedom (df) for a paired samples t-test is equal to the number of pairs minus 1. In this case, there are 8 pairs, so the df is 7.

To find the t-critical value, we need to look up the critical value for a one-tailed t-test with a significance level of 0.05 and df = 7. By consulting a t-table, we find that the critical value is approximately 1.895.

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

t-obtained = (Dbar - 0) / (Sd / sqrt(n))

where Dbar is the mean difference (8.04), Sd is the standard deviation (sqrt(S2D), which is sqrt(79.22)), and n is the number of pairs (8).

Plugging in the values, we get:

t-obtained = (8.04 - 0) / (sqrt(79.22) / sqrt(8))

After evaluating the expression, we find that the t-obtained value is approximately 5.09.

The conclusion in APA format with statistical evidence is as follows:

The t-obtained value of 5.09 is greater than the t-critical value of 1.895, with 7 degrees of freedom and a significance level of 0.05. Therefore, we reject the null hypothesis. This means that there is a significant difference in the growth of plants when using different fertilizers. Plants given fertilizer A (mean = 13 cm) grew significantly more than plants given fertilizer B (mean = 8 cm).

To know more about Degrees of freedom (df), visit:

https://brainly.com/question/28270067

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