![]() However, Python 2.6+ and 3.x offer a mutable byte string as bytearray. ![]() Mostly, you should use it when you need to expose a C array to an extension or a system call (for example, ioctl or fctnl).Īrray.array is also a reasonable way to represent a mutable string in Python 2.x ( array('B', bytes)). It can hold only homogeneous data (that is to say, all of the same type) and so it uses only sizeof(one object) * length bytes of memory. The array.array type, on the other hand, is just a thin wrapper on C arrays. But they use a lot more space than C arrays, in part because each item in the list requires the construction of an individual Python object, even for data that could be represented with simple C types (e.g. If you need to shrink and grow your list time-efficiently and without hassle, they are the way to go. Print("the value that separates 0.Basically, Python lists are very flexible and can hold completely heterogeneous, arbitrary data, and they can be appended to very efficiently, in amortized constant time. #the value that separates 0.05 (thus, 5%) of the data from 95% #The p value obtained from the one sample t-test is significant Print("t-value=", tval, "p-value=", pval) Tval = (mean_after_med-mean_before_med)/s #Numpy's var function uses n as divisor as a default but when ddof is set to 1, it uses n-1 as divisor Print('Mean after medication:', mean_after_med) Print('Mean before medication:', mean_before_med) ![]() #H1: The medication lowers blood pressureīefore_med = np.array()Īfter_med = np.array() This is my code (where I use var_med = np.var(med_diff, ddof=1) and s = np.sqrt(var_med/n): #H0: The medication has no effect on blood pressure Could somebody confirm this please?ĭisclaimer: I know that there are different modules to do a t-test or to calculate the std with numpy but the prof wants it somehow like this, Computer Linguists, sikes. However, I thought just taking the square root of the variance is enough, without dividing by n again. ![]() Presuming that var_med = np.var(med_diff, ddof=1) is correct, do I need to divide the variance by n again under the square root? In my case, the prof has written it like this: s = np.sqrt(var_med/n).Calculating var_med with np.var(med_diff, ddof=1), does it imply that med_diff is divided by n-1 (bcs of ddof=1) by numpy itself? Albeit the confusing documentation on this, I do believe it does that afaik (this is also what I assume for the third question).Is it right that I calculate the variance with the difference between both arrays (because my intuition tells me that I should use the mean difference, ergo mean_after_med - mean_before_med)?.Conduct a one-sided t-test to determine if the medication isĮffective at lowering blood pressure. The alternative hypothesis is that the medication lowers blood pressure. Medication has no effect on blood pressure. Want to know if the medication lowers blood pressure. Measure the blood pressure of 20 participants before and after taking the medication, and Suppose a researcher is studying the effect of a new medication on blood pressure. To calculate the variance and the means, then compare them according to the t-test. For both of these exercise, you get two lists of values that you want to compare.
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