With the help of t test find if significant difference exist between the scores obtained on emotional intelligence by male and female teachers

Question

Male teachers - 45, 32, 25, 57, 36, 42, 35, 55, 66, 65, 30, 35, 22, 27, 26

Female teachers - 36, 53, 64, 55, 52, 34, 62, 73, 61, 34, 45, 38, 36, 25, 45

Answer 1

The formulas for above re on the nist,gov website. This is a two tailed test and the value to the left of -.73919 with 28 degrees of freedom is over 46 % which isn't going to pass any meaningful test.

µso ACCEPT H0: that µ1 = µ2 and reject H1: µ1 ≠µ2

Answer 2

The null hypothesis is that the means are equal.

The alternate hypothesis is that the means are not equal.

You will need to compute the means and standard deviations of the individual samples.

Sample size men (n1) = women (n2) = 15

The t-statistic to be used is (mean male - mean female)/sqrt[std dev men^2/n1^2 + std dev women^2/n2^2)]

The degrees of freedom is smaller of n1 - 1 and n2 - 1, which is 14

Find the p-value corresponding to the computed value of the t-statistic with 14 degrees of freedom, remembering that is a 2-sided test, so p-value is double each tail value.

p-value is the smallest level of significance at which you can reject the null hypothesis, e.g., if p = 0.045, then you can reject the null hypothesis 0.05 level of significance.

Answer 3

To get the test statistic, t, you need 6 things:

1. n₁, the sample size of the first group
2. n₂, the sample size of the second group
3. xbar₁, the sample mean of the first group
4. xbar₂, the sample mean of the second group
5. s₁, the sample standard deviation of the first group
6. s₂, the sample standard deviation of the second group

The sample sizes are the easiest, just count how many observations are in each group. They both have 15 observations, so n₁ = 15 and n₂ = 15.

To get each sample mean, you simply add up all the values in that group and divide by the sample size. After some work, you find that xbar₁ = 39.86667 and xbar₂ = 47.53333.

Sample standard deviation is a little more complicated. I'm not sure if your instructor wants you to do it by hand, or with some software. But if you have any kind of graphing calculator, you should be able to input a list of values and get the sample mean and sample standard deviation. (There will also be a way to so this t-test automatically, but here I'm showing you how to do it "by hand"). The formula for sample standard deviation is Σ(x_i-xbar)²/(n-1). You take each observation, you subtract the sample mean, you square the difference, you add up all those squared differences and then you divide the sum total by the sample size minus 1. After some work, you find that s1 = 14.60854 and s2 = 13.78336. (Hint, you can use wolframalpha.com for stuff like this. Just type "sd()" and in the parentheses put a list of numbers).

Now we have all 6 things we need, we just need to put them together into the formula for a t-statistic:

t = (xbar₁-xbar₂)/√(s₁²/n₁ + s₂²/n₂)

t = (39.86667-47.53333)/√(14.60854²/15 + 13.78336²/15) = -1.47839

Since the problem tells us we're looking for a significant difference, not that mean 1 is greater than mean 2, or that mean 2 is greater than mean 1, we want to perform a two tailed test. We can use a table such as this one: https://www.medcalc.org/manual/t-distribution.php.

Since our test statistic is negative, and this table has no negative values, and the t-distribution is symmetric, we can simply look at the p-value for 1.478 rather than -1.478. But first, we need to determine what the degrees of freedom are. In a two sample t-test, the degrees of freedom is n₁ + n₂ - 2 = 15 + 15 - 2 = 28.

So on the left side of the table, lets scroll down to 28, and then lets move over to the right until we find a value close to 1.478. We see that it's somewhere between the first and second values on the table, which, by looking at the top of the columns we see is associated with p-values of 0.2 and 0.1. So we can say that our p-value is somewhere between 0.1 and 0.2, which is larger than the commonly used significance level of 0.05. Our p-value is large enough that we can say there is insufficient evidence to conclude that the mean emotional intelligence is significantly different between male and female teachers. (i.e. we fail to reject H₀).