Question

The average undergraduate cost for tuition, fees, and room and board for two-year institutions last year was $13,252. The following year, a random sample of 20 two-year institutions had a mean of $15,560 and a standard deviation of $3500. Is there sufficient evidence at the ?= 0.05 level to conclude that the mean cost has increased. Solve the question by traditional approach.

Question no 2: A large university reports that the mean salary of parents of an entering class is $91,600. To see how this compares to his university, president surveys 28 randomly selected families and finds that their average income is $88,500. If the standard deviation is $10,000, can the president conclude that there is a difference? At ?= 0.10, is he correct? Test the hypothesis by P-value approach.

Question no 3: The manager of a large company claims that the standard deviation of the time (in minutes) that it takes a telephone call to be transferred to the correct office in her company is 1.2 minutes or less. A sample of 15 calls is selected, and the calls are timed. The standard deviation of the sample is 1.8 minutes. At ?= 0.01, test the claim that the standard deviation is less than 1.2 minutes. Use the P-value method

Answer 1

ANSWER:

Traditional method:

1) n = 20 , mean = $15,560 , s = $3,500

alpha = 0.05 , d.f. = n - 1 = 20 - 1 = 19

h0 : u = 13,252

h0 : u > 13,252

t = ( ( mean - u) / (s / sqrt(n) ) )

t = (15560 - 13252) / ( 3500 / sqrt(20) )

t = 2308 / ( 3500 / 4.47)

t = 2308 / 782.62

t = 2.94

t value of alpha = 0.05 is 1.729

Reject the null hypothesis since the test value falls within the critical region and there is sufficient evidence to suggest that the average tution cost has increased.

2) p value method:

h0 : u = 91,600

h0 : u not equal to 91,600.

n = 28 , mean = 88,500 , u = 91,600 , df = n - 1 = 28 - 1 = 27 , s = 10,000

t statistic = ( ( mean - u ) /  / (s / sqrt(n) ) )

t statistic = ( 88500 - 91600) / ( 10000 / sqrt (28)

t statistic = ( -3100) / ( 10000 / 5.29)

t statistic = (-3100 / 1889.82)

t statistic = -1.64

p value of t statistic is 0.1126

p value is greater then the alpha 0.1 that 0.1126 > 0.1 , therefore the null hypothesis is accepted.

3) h0 : sigma <= 1.2

h1 : sigma > 1.2

alpha = 0.01 , s = 1.8 , n = 15 , df = n - 1 = 15 - 1 =14

x^2 = ( n - 1) * s^2 / ( sigma) ^ 2

x^2 = ( 15 - 1) * (1.8) ^ 2 / ( 1.2) ^ 2

x^2 = 14 * 3.24 / (1.44)

x^2 = 45.36 / 1.44 = 31.5

p value of 31.5 is 0.0047 which is less then 0.005 (two tailed test) and that is why we reject the null hypothesis and there is not enough evidencethat the standard deviation is less than or equal to 1.2 minutes.