Question

A company sent its employees to attend two different English courses. The company is interested in knowing if there is any difference between the two courses attended by its employees. The company asked them to take a common test. The summary statistics of the test results of each of the two English courses are recorded in the following table: Course 1 93 99 105 94 82 70 86 Course 2 118 94 106 72 90 66 153 98 a. Assume the two population variances are equal, is there enough evidence to conclude that there are differences between the two courses at 0.1 significance level? b. The Labour Department reports that historically the mean deductible amount per employee is $502 with a standard deviation of $100. i. If we select a random sample of 60 employees’ records, compute the standard error of the sample mean for the consultancy company’s survey. ii. What is the chance the consultancy company finds a sample mean between $477 and $527? iii. What is the probability the sample mean is greater than $530?

Answer 1

Let X: Course 1 data and Y: Course 2 data

We have to calculate sample means ( & ), sample standard deviations (S1 & S2) and Sample sizes (n1 & n2) for both X and Y data, we get

For X: = 89.86, S1 = 11.625, n1 = 7

For Y: = 99.63, S2 = 27.381, n2 = 8

Step I: Set up the null and alternate hypothesis

(There is no difference between two courses)

(There is difference between two courses)

Step II: Determine test statistic, assuming population variances are equal

Where, Sp is pooled sample standard deviation and calculated by using below formula

  

Putting this value in equation (I), we get

Step III: Calculate degrees of freedom (df):

df = n1 + n2 - 2 = 7 + 8 - 2 = 13

Step IV: for = 0.1, df = 13, critical t value is +/- 1.771

Step V: Calculate p-value:

p = p(t -0.8742) = 0.3979

Step VI: Decision & Conclusion: Since = 0.10 < p = 0.3979, we failed to reject the null hypothesis and conclude that there is no mean difference between two courses at 0.1 level of significance.

b. The Labour Department reports that historically the mean deductible amount per employee is $502 with a standard deviation of $100

i.) If we select a random sample of 60 employees’ records, compute the standard error of the sample mean for the consultancy company’s survey.

We have given,

= 502, = 100, n = 60

We are asked to find standard error of sample mean (SE)

ii.) What is the chance the consultancy company finds a sample mean between $477 and $527?

That is

= p(Z < 1.94) - p(Z < -1.94)
= NORMSDIST(1.94) - NORMSDIST(-1.94)

= 0.9736 - 0.0264

= 0.9472

iii.) What is the probability the sample mean is greater than $530?

that is,

= 1 - p(Z < 2.17)

= 1 - NORMSDIST(2.17)

= 1 - 0.9850

= 0.0150