Question

1.

Toll Brothers is a luxury home builder that would like to test the hypothesis that the average size of new homes exceeds 2,400 square feet. A random sample of 36 newly constructed homes had an average of 2,510 square feet. Assume that the standard deviation of the size for all newly constructed homes is 480 square feet. Toll Brothers would like to set α = 0.02. Use the p-value approach to test this hypothesis.

	a.	0.0838
	b.	0.01
	c.	0.05
	d.	0.1 2. Bananas are sold in bunches at a grocery store and typically consist of 4-8 bananas per bunch. Suppose the weight of these bunches follows a normal distribution with a mean of 3.54 pounds and a standard deviation of 0.63 pounds. The interval around the mean that contains 99.7% of the bunch weights is ________. 1.02, 6.06 0.39, 6.69 1.65, 5.43 2.28, 4.80 3. Bob's golf score at his local course follows the normal distribution with a mean of 92.1 and a standard deviation of 3.8. The interval around the mean that contains 68% of Bob's golf scores is ________. 90.2, 94.0 80.7, 103.5 84.5, 99.7 88.3, 95.9 4. Costco installs automobile tires on a first-come first-serve basis. The total time a customer needs to wait for the installation to be completed follows the normal distribution with a mean time of 106.3 minutes and a standard deviation of 18.5 minutes. What is the probability that a randomly selected customer will wait between 80 and 95 minutes for his or her tires to be installed? 0.2960 0.1245 0.3455 0.1931 5. Bananas are sold in bunches at a grocery store and typically consist of 4-8 bananas per bunch. Suppose the weight of these bunches follows a normal distribution with a mean of 3.54 pounds and a standard deviation of 0.63 pounds. What is the probability that a randomly selected bunch of bananas will weigh more than 3.0 pounds? 0.9463 0.5398 0.6950 0.8051 6. YouTube would like to test the hypothesis that the average length of an online video watched by a user is more than 8 minutes. A random sample of 37 people watched online videos that averaged 8.7 minutes in length. It is believed that the population standard deviation for the length of online videos is 2.5 minutes. YouTube would like to set α = 0.02. The correct hypothesis statement for this hypothesis test would be _________________________. H0: μ ≥ 8.0; H1: μ < 8.0 H0: μ ≠ 8.0; H1: μ = 8.0 H0: μ ≤ 8.0; H1: μ > 8.0 H0: μ = 8.0; H1: μ ≠ 8.0 7. According to College Board, the average room and board costs for a private four-year college for the 2011-2012 academic year was $10,090. Assume that this cost follows the normal distribution with a standard deviation of $1,125. Determine the interval of age around the mean that includes 95% of the sample mean

Answer 1

1)

Hypothesis : ,     Right tailed test.

Given, = 2510 , = 480 , n = 36

Test statistic :

p-value :

P-value for this right tailed test is given by,

p-value = P( z > 1.38 )   = 1 - P( z < 1.38 )

Using z score table,

P( z < 1.38 ) = 0.9162

So, P( z > 1.38 ) = 1 - 0.9162 = 0.0838

P-value = 0.0838

2)

Given,

weight of these bunches follows a normal distribution with a mean of = 3.54 pounds and a standard deviation of = 0.63

According to Empirical rule of normal distribution, Approximately 99.7% of the data falls within three standard deviations of the mean. The mathematical notation for this is:

So,The interval around the mean that contains 99.7% of the bunch weights is ( 3.54 -3*0.63 , 3.54 +3*0.63)

The interval around the mean that contains 99.7% of the bunch weights is ( 1.65 , 5.43 )

3)

Given,

Bob's golf score at his local course follows the normal distribution with a mean of = 92.1 and a standard deviation of = 3.8

According to Empirical rule, Approximately 95% of the data falls within two standard deviations of the mean. The mathematical notation for this is:

So,The interval around the mean that contains 68% of Bob's golf scores is ( 92.1 - 2*3.8 , 92.1 + 2*3.8 )

The interval around the mean that contains 68% of Bob's golf scores is ( 84.5 , 99.7 )

4)

Let , X be the total time a customer needs to wait for the installation to be completed.

X follows normal distribution with = 106.3 and = 18.5

We have to find P( 80 < x < 95 )

P( 80 < x < 95 ) = P( x < 95 ) - P(x < 80 )

Using Excel function ,   =NORMDIST( x , , , 1 )

P( x < 95 ) =NORMDIST( 95 , 106.3 , 18.5 , 1 ) = 0.2707

p( x < 80 ) = NORMDIST( 80 , 106.3 , 18.5 , 1 ) = 0.0776

So, P( 80 < x < 95 ) = 0.2707 - 0.0776 = 0.1931

The probability that a randomly selected customer will wait between 80 and 95 minutes for his or her tires to be installed is 0.1931