2. For the Salary _ Experience data set, test to see if there is a difference in
ID: 3225395 • Letter: 2
Question
2. For the Salary _ Experience data set, test to see if there is a difference in salaries for employees who selection into their current position was from within the company (internal employees) versus who were selected into their position from outside the company (external employees).
The XL output is as follows. Comparing Salaries (x$1000)
t-test : two-sample Assuming non-equal variances
GIVE TYPED ANSWER NOT HAND WRITTEN
External salary A Internal salary B mean 66.90 78.35 Ivarianci 87.04 79.68 02 88 observations Pooled variance 83.08 Hypothesized MeanDifference 148 df T stat 7.581 POT -t) one- tail 0.000 t-critical one -tail 1.655 0.000 POT -t) two-tail t-critical one -tail 1.075 a. State the null hypotheses and alternative hypotheses. b. What is test statistics give answer with formula. c. ConclusionExplanation / Answer
Solution:-
State the hypotheses. The first step is to state the null hypothesis and an alternative hypothesis.
Null hypothesis: 1 - 2 = 0
Alternative hypothesis: 1 - 2 0
Note that these hypotheses constitute a two-tailed test. The null hypothesis will be rejected if the difference between sample means is too big or if it is too small.
Formulate an analysis plan. For this analysis, the significance level is 0.10. Using sample data, we will conduct a two-sample t-test of the null hypothesis.
Analyze sample data. Using sample data, we compute the standard error (SE), degrees of freedom (DF), and the t statistic test statistic (t).
SE = sqrt[(s12/n1) + (s22/n2)]
SE = 1.524
DF = 148
t = [ (x1 - x2) - d ] / SE
t = 7.51
where s1 is the standard deviation of sample 1, s2 is the standard deviation of sample 2, n1 is the size of sample 1, n2 is the size of sample 2, x1 is the mean of sample 1, x2 is the mean of sample 2, d is the hypothesized difference between the population means, and SE is the standard error.
Since we have a two-tailed test, the P-value is the probability that a t statistic having 148 degrees of freedom is more extreme than -1.524; that is, less than -1.524 or greater than 1.524.
Thus, the P-value = less than 0.00001
Interpret results. Since the P-value (almost 0) is less than the significance level (0.10), we cannot accept the null hypothesis.
From the above test we have sufficient evidence in the favor of thr claim that there is significant difference between external and internal salary.
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