1. An experiment was conducted to compare plant growth by treating surrounding p
ID: 3375780 • Letter: 1
Question
1.
An experiment was conducted to compare plant growth by treating surrounding plant soil with one of two fertilizers. Fifty of 150 seedlings were randomly chosen to receive Fertilizer X. The rest received Fertilizer Y. The heights were recorded for each seedling, in centimeters, at the end of the study. The results obtained are shown.
Fertilizer X: X?1 = 12.4 cm, s1 = 4 cm
Fertilizer Y: X?2 = 18.7 cm, s2 = 3 cm
We wish to determine whether there is a difference in height for the seedlings treated with the different fertilizers by testing the hypotheses H0: ?2 ? ?1 = 0 and Ha: ?2 ? ?1 ? 0. Find the 90% confidence interval for ?2 ? ?1 and make a conclusion based on that confidence interval.
The 90% confidence interval is 6.3 ± 7.37. The p-value is less than 0.10.
The 90% confidence interval is 6.3 ± 1.05. I would not reject the null hypothesis of no difference at the 0.10 level.
The 90% confidence interval is 6.3 ± 5.23. I would not reject the null hypothesis of no difference at the 0.10 level.
The 90% confidence interval is 6.3 ± 5.23. I would reject the null hypothesis of no difference at the 0.10 level.
The 90% confidence interval is 6.3 ± 1.05. I would reject the null hypothesis of no difference at the 0.10 level.
2.
In conducting a test on the hypotheses H0: µ = 250 and Ha: µ < 250, you find that the population mean is 250 when it is actually 247. This results in what type of error?
No error
Type II error
Type I error
Standard deviation of the mean
There is not enough information given
3.
The assumption of the Law of Averages has been violated.
The variables X and Y are not related at all.
A line is an appropriate model to describe the relation between X and Y.
A line is not an appropriate model to describe the relation between X and Y.
There is not enough information about the variables X and Y to form a conclusion.
Explanation / Answer
Solution
(1) ?2 ? ?1 = 18.7-12.4 =6.3
Standard error = (16/50 + 9/100)0.5 = 0.6403
margin of error = 0.6403*1.645 = 1.05
The 90% confidence interval is 6.3 ± 1.05. I would reject the null hypothesis of no difference at the 0.10 level.
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