High School

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 that 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, \(\bar{D} = 8.04\) and the variance is \(S^2_D = 79.22\). Plants given fertilizer A grew to 13 cm while plants given fertilizer B grew to 8 cm. Use an alpha of .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?
4. What is your t-obtained?
5. What is your conclusion in APA format with statistical evidence? Do you reject or fail to reject the null hypothesis?

Answer :

This is a paired t-test because the plants are paired and each pair is given a different fertilizer.

The degrees of freedom for a paired t-test are equal to the number of pairs minus 1, which in this case is 8 - 1 = 7.

With an alpha level of 0.05 and 7 degrees of freedom, we can look up the t-critical value from the t-distribution table. The t-critical value for a two-tailed test with 7 degrees of freedom at alpha = 0.05 is approximately ±2.364.

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

t = (Dbar - μD) / (Sd / √n)

where Dbar is the mean difference, μD is the hypothesized mean difference (which is 0 in this case), Sd is the standard deviation of the differences, and n is the number of pairs.

In this case, Dbar = 8.04, μD = 0, Sd = √S2D = √79.22, and n = 8.

t = (8.04 - 0) / (√79.22 / √8) ≈ 8.04 / 4.454 ≈ 1.806

The t-obtained value is approximately 1.806.

To interpret the results, we compare the t-obtained value to the t-critical value. If the t-obtained value falls within the t-critical range, we fail to reject the null hypothesis. If the t-obtained value is outside the t-critical range, we reject the null hypothesis.

In this case, the absolute value of the t-obtained value (1.806) is smaller than the t-critical value (2.364). Therefore, we fail to reject the null hypothesis. This means that there is not enough evidence to conclude that there is a significant difference in plant growth between the two fertilizers.

To learn more about range click here:

brainly.com/question/29204101

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