Two Independent simple random samples are taken to test the difference between the means of two populations. The standard deviations are not known, but are assumed to be equal. The sample sizes are n1 = 15 and n2 = 35. The correct distribution to use is the t distribution with 51 degrees of freedom z distribution with 50 degrees of freedom z distribution with 49 degrees of freedom t distribution with 48 degrees of freedom

The t test for the difference between the means of two independent populationsalways assumes that the respectivea. sample sizes are equalb. sample variances are equalc. populations are normally distributed if each sample size is less than 30d. All of the above

If we have a sample size of 15 and the population standard deviation is known, we will use: t- test for hypothesis testing z-test for hypothesis testing both t and z test F Test

If two populations are independent and each normally distributed and the population standard deviations are not equal, the test statisticT=X¯1−X¯2−(μ1−μ2)S21n1+S22n2‾‾‾‾‾‾‾‾√𝑇=𝑋¯1−𝑋¯2−(𝜇1−𝜇2)𝑆12𝑛1+𝑆22𝑛2 follows a:Question 4Select one:a.t-distribution with n1+n2−2 degrees of freedomt-distribution with 𝑛1+𝑛2−2 degrees of freedom b.standard normal distribution with n1+n2−2 degrees of freedomstandard normal distribution with 𝑛1+𝑛2−2 degrees of freedom c.t-distribution with degrees of freedom equal to (S21n1+S22n2)2(S21/n1)2n1−1+(S22/n2)2n2−1t-distribution with degrees of freedom equal to (𝑆12𝑛1+𝑆22𝑛2)2(𝑆12/𝑛1)2𝑛1−1+(𝑆22/𝑛2)2𝑛2−1 d.t-distribution with n−1 degrees of freedom

The sampling distribution of two populations  is approximated by a t-distribution with n1 + n2 degrees of freedom normal distribution t-distribution with n1 + n2 - 1 degrees of freedom t-distribution with n1 + n2 + 2 degrees of freedom

The statistical distribution used for testing the difference between two populationvariances is thea. t distributionb. binomial distributionc. standardised normal distributiond. F distribution