A farmer wants to see if a new kind of fertilizer can help him grow better fruits. He prepares two separate plots of land and adds the fertilizer in one of those only. After a month he measure the weights of the fruits in each plot. Here are the results: plot 1 (no fertilizer) plot2 (fertilizer) sample mean Ino fert. = 3.5 Tfert. = 3.8 %3D sample standard deviation Sno fert. = 0.8 Sfert. 0.9 sample size Nno fert. = 120 Nfert. 110 Should this farmer invest in fertilizers? That is, is the mean weight of fruits significantly higher for the plot with fertilizer compared to the plot without it at the a = 0.1 level. What is the critical value for this test? O qt(0.95, df = 109) = 1.658953 O qt(0.9, df = 228) = 1.285276 O qt(0.9, df = 119) = 1.288706 O qt(0.9, df = 109) = 1.289367 %3D
Inverse Normal Distribution
The method used for finding the corresponding z-critical value in a normal distribution using the known probability is said to be an inverse normal distribution. The inverse normal distribution is a continuous probability distribution with a family of two parameters.
Mean, Median, Mode
It is a descriptive summary of a data set. It can be defined by using some of the measures. The central tendencies do not provide information regarding individual data from the dataset. However, they give a summary of the data set. The central tendency or measure of central tendency is a central or typical value for a probability distribution.
Z-Scores
A z-score is a unit of measurement used in statistics to describe the position of a raw score in terms of its distance from the mean, measured with reference to standard deviation from the mean. Z-scores are useful in statistics because they allow comparison between two scores that belong to different normal distributions.

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