please answer the question fully! You might think that increasing the nutrient resources available would elevate the number of plant species that an area couldsupport, but the evidence suggests otherwise. The data in the accompanying table are from the Park Grass Experiment atRothamsted Experimental Station in the U.K., where grassland field plots have been fertilized annually for the past 150 years(collated by Harpole and Tilman 2007). The number of plant species recorded in 10 plots is given in response to the numberof different nutrient types added in the fertilizer treatment (nutrient types include nitrogen, phosphorus, potassium, andso on).Plot1234678910Number ofnutrients added0001231344Number ofplant species36363234333020232116Download the data in your preferred format.CSV Excel JMP Mac-Text Minitab PC-Text R SPSS TI CrunchItTest the null hypothesis of no treatment effect on the number of plant species. Determine the value of the test statistic t.Give your answer to two decimal places.

To test the null hypothesis of no treatment effect on the number of plant species, we can use a statistical test such as the Pearson correlation coefficient. The null hypothesis (H0) is typically that there is no relationship between the two variables, in this case, the number of nutrients added and the number of plant species. The alternative hypothesis (H1) would then suggest that there is a significant relationship.Let us denote:x as the number of nutrients added.y as the number of plant species.The null hypothesis (H0) and alternative hypothesis (H1) can be stated as follows:H0:There is no correlation between the number of nutrients added (x) and the number of plant species (y). AgainstH1:There is a correlation between the number of nutrients added (x) and the number of plant species (y).The correlation coefficient (r) can be calculated using the formula:r=∑(xi−x¯)(yi−y¯)∑(xi−x¯)2∑(yi−y¯)2where:xi and yi are the individual data points.x¯ and y¯ are the means of x and y, respectively.The pearson correlation coefficient obtained between the two variables is obtained as r=−0.732The test statistic is obtained as:t=rn−21−r2t=−0.73214−21−0.535824t=−2.53572240.464176t=−2.53572240.6813046308t=−3.7218628575t=−3.72The test statistic is obtained as -3.72The degrees of freedom is obtained as:df=n−2=14−2=12The p-value of the test is obtained as: 0.016Here we will consider the standard significance level of 0.05.Conclusion: Since the p-value is less than the significance level, we will reject the null hypothesis and conclude on the basis of the given data that there is a significant relationship between the number of nutrients added (x) and the number of plant species (y).