Question

1. Mean entry-level salaries for college graduates with mechanical engineering degrees and electrical engineering degrees are believed to be approximately the same. A recruiting office thinks that the mean mechanical engineering salary is actually lower than the mean electrical engineering salary. The recruiting office randomly surveys 44 entry level mechanical engineers and 58 entry level electrical engineers. Their mean salaries were $46,300 and $46,900, respectively. Their standard deviations were $3420 and $4230, respectively. Conduct a hypothesis test at the 5% level to determine if you agree that the mean entry- level mechanical engineering salary is lower than the mean entry-level electrical engineering salary. Let the subscript m = mechanical and e = electrical.
NOTE: If you are using a Student's t-distribution for the problem, including for paired data, you may assume that the underlying population is normally distributed. (In general, you must first prove that assumption, though.)

Part (d)State the distribution to use for the test. (Enter your answer in the form z or t_df where df is the degrees of freedom. Round your answer to two decimal places.)

Part (f)What is the p-value? (Round your answer to four decimal places.)

2.Some manufacturers claim that non-hybrid sedan cars have a lower mean miles-per-gallon (mpg) than hybrid ones. Suppose that consumers test 21 hybrid sedans and get a mean of 31 mpg with a standard deviation of 7 mpg. Thirty-one non-hybrid sedans get a mean of 20 mpg with a standard deviation of three mpg. Suppose that the population standard deviations are known to be six and three, respectively. Conduct a hypothesis test at the 5% level to evaluate the manufacturers claim.

Part (d)State the distribution to use for the test. (Round your answers to two decimal places.)

Part (e)What is the test statistic? (If using the z distribution round your answer to two decimal places, and if using the t distribution round your answer to three decimal places.)

Part (f)What is the p-value? (Round your answer to four decimal places.)

NOTE: If you are using a Student's t-distribution for the problem, including for paired data, you may assume that the underlying population is normally distributed. (In general, you must first prove that assumption, though.)

Answer 1

(d)

Since we are comparing the mean scores of two samples and we do not know the population standard deviations, we will use two-sample t test.

We need to check for homogeneity of variances.

Test statistic, F = s2^2 / s1^2   = 4230^2 / 3420^2  = 1.53

Numerator df = n2 - 1 = 58 - 1 = 57

Denominator df = n1 - 1 = 44 - 1 = 43

Critical value of F at 0.05 significance level and df = 57, 43 is 1.62

Since the observed F (1.53) is less than the critical value, we conclude that there is no significant evidence that the population variance (or standard deviations) of the two groups are not equal.

Assuming equality of variance, df = n1 + n2 - 2 = 44 + 58 - 2 = 100

The distribution to use for the test is t distribution with df = 100 or,

(f)

So, we assume that the population standard deviations of both groups to be equal and we can apply pooled t-test.

Pooled variance Sp^2 = [(n1 - 1) s1^2   + (n2 - 1)s2^2 ] / (n1 + n2 - 2)

= [(44 - 1) 3420^2 + (58 - 1) ] 4230^2 / (44 + 58 -2)

= 15228405

Standard error of mean difference = sqrt( Sp^2 * ((1/n1) + (1/n2)))

= sqrt( 15228405* ((1/44) + (1/58)))

= 780.1659

Test statistic, t = (x1 - x2) / Std error

= (46300 - 46900) / 780.1659

= -0.769

P-value = P(t < -0.769, df = 100) = 0.2219

(d)

Since we are comparing the mean scores of two samples and we do not know the population standard deviations, we will use two-sample t test.

We need to check for homogeneity of variances.

Test statistic, F = s2^2 / s1^2   = 7^2 / 3^2  = 5.44

Numerator df = n1 - 1 = 21 - 1 = 20

Denominator df = n2 - 1 = 31 - 1 = 30

Critical value of F at 0.05 significance level and df = 20, 30 is 1.93

Since the observed F (5.44) is gretaer than the critical value, we conclude that there is significant evidence that the population variance (or standard deviations) of the two groups are not equal.

As, we cannot assume that the population standard deviations of two groups are equal, we use welch method to compute degree of freedom.

DF = (s1^2/n1 + s2^2/n2)^2 / { [ (s1^2 / n1)^2 / (n1 - 1) ] + [ (s2^2 / n2)^2 / (n2 - 1) ] }

= (7^2/21 + 3^2/31)^2 / { [ (7^2 / 21)^2 / (21 - 1) ] + [ (3^2 /31)^2 / (31 - 1) ] }

25 (Rounding to nearest integer)

The distribution to use for the test is t distribution with df = 25 or,

(e)

Standard error of mean difference = sqrt[ (s1^2/n1) + (s2^2/n2) ]

= sqrt[ (7^2/21) + (3^2/31) ]

= 1.61977

Test statistic, t = (x1 - x2) / Std error

= (31 - 20) / 1.61977

= 6.791

(f)

P-value = P(t > 6.791, df = 25) = 0.0000