Throughout the problem we assume that p, and with it ˆp, is constant. a) What effect has an increase of n on the standard error? b) By what factor is the standard error changing if the sample size is changed from n to 2n? c) When changing the sample size from n to kn, what value of k reduces the standard error by a factor of 1/2?

I need assistance with D and E Only.  Given the sample data. x: 23, 17, 15, 32, 27 (a) Find the range. (Enter an exact number.) (b) Verify that Σx = 114 and Σx2 = 2,796. (For each answer, enter an exact number.)Σx = Σx2 = (c) Use the results of part (b) and appropriate computation formulas to compute the sample variance s2 and sample standard deviation s. (For each answer, enter a number. Round your answers to two decimal places.)s2 = s = (d) Use the defining formulas to compute the sample variance s2 and sample standard deviation s. (For each answer, enter a number. Round your answers to two decimal places.)s2 = s = (e) Suppose the given data comprise the entire population of all x values. Compute the population variance σ2 and population standard deviation σ. (For each answer, enter a number. Round your answers to two decimal places.)σ2 = σ =

The degrees of freedom is critical to computing the p-value for the chi-square goodness of fit. We will discuss the p-value in the next lab. The degrees of freedom for the Chi-Square goodness of fit is equal to (the number of non-zero rows in the table) - 1. The number of non-zero rows is the number of expected values that are non-zero. For example: Phenotype ... Expected ... Disease ... 234 ... Wild-type ... 6743 ... has two rows where the Expected value is non-zero. So the degrees of freedom is 2-1 = 1. Another example: Phenotype ... Expected ... Disease ... 2374 ... Wild-type ... 0 ... has one row where the Expected value is non-zero. So the degrees of freedom is 1-1 = 0. With 0 degrees of freedom, the chi-square test cannot be calculated. A third example (sex-linked): Phenotype ... Expected ... Disease-Male ... 2374 ... Disease-Female ... 9456   Wild-type-Male ... 1001 ... Wild-type-Female ... 235 ... has four rows where the Expected values are…

Assume that the population proportion is 0.46. Compute the standard error of the proportion,  σp,  for sample sizes of 500,000; 1,000,000; 5,000,000; 10,000,000; and 100,000,000. (Round your answers to five decimal places.) sample size of 500,000sample size of 1,000,000sample size of 5,000,000sample size of 10,000,000sample size of 100,000,000 What can you say about the size of the standard error of the sample proportion as the sample size is increased? The standard error of the sample proportion,  σp,   ---Select--- increases decreases and becomes extremely  ---Select--- small large as the sample size becomes huge.