Question

The management of White Industries is considering a new method of assembling its golf cart. The present method requires 62.3 minutes, on the average, to assemble a cart. The mean assembly time for a random sample of 20 carts, using the new method, was 60.6 minutes, and the standard deviation of the sample was 3.1 minutes. Using the 0.02 level of significance, can we conclude that the assembly time using the new method is faster?

a. What is the decision rule? (Negative answer should be indicated by a minus sign. Round the final answer to 3 decimal places.) Reject H0: μ ≥ 62.3 and accept H1: μ < 62.3 when the test statistic is less than:____

b. What is the value of the test statistic? (Negative answer should be indicated by a minus sign. Round the final answer to 3 decimal places.)

Answer #1

Step 1 :

Ho:

Ha:

Step 2

Assuming that the data is normally distributed. Since the population standard deviation is not given we will use t statistics.

n= 20

sample mean = 60.6

sample standard deviation = 3.1

Step 3

level of significance = 0.02

The t-critical value for a left-tailed test, for a significance level of α=0.02

tc = −2.205

Since the t stat ( -2.452) is less than t critical we reject the Null hypothesis. [i.e. t stat is in the rejection area]

ANS

a) Reject H0: μ ≥ 62.3 and accept H1: μ < 62.3 when the test
statistic is less than : **-2.205**

b) t statistics = **-2.452**

The management of White Industries is considering a new method
of assembling its golf cart. The present method requires 50.3
minutes, on the average, to assemble a cart. The mean assembly time
for a random sample of 60 carts, using the new method, was 48.6
minutes, and the standard deviation of the sample was 2.8
minutes.
Using the 0.10 level of significance, can we conclude that the
assembly time using the new method is faster?
a. What is the decision...

The management of White Industries is considering a new method
of assembling its golf carts. The present method requires an
average of 42.3 minutes to assemble a cart. The mean assembly time
for a random sample of 25 carts, using the new method, was 41.0
minutes. From historical information, the population standard
deviation is known to be 2.7 minutes. Using an alpha value of .10,
can we conclude that the assembly time using the new method is
different than it...

The operations manager of Slug Computers is considering a new
method of assembling its Mollusk Computer. The present method
requires 42.3 minutes, on average to assemble the speedy computer.
The new method was introduced and a random sample of 24 computers
revealed a mean assembly time of 40.6 minutes with a standard
deviation of 2.7 minutes, the distribution is normal.
At the .10 level of significance, can we conclude the new method
is faster, on average?
Estimate the p value....

Moen has instituted a new manufacturing process for assembling a
component. Using the old process, the average assembly time was
4.25 minutes per component. After the new process was in place for
30 days, the quality control engineer took a random sample of
fourteen components and noted the time it took to assemble each
component. Based on the sample data, the average assembly time was
4.38 minutes, with a standard deviation of .16 minutes. Using a
significance level of .02,...

Chicken Delight claims that 92% of its orders are delivered
within 10 minutes of the time the order is placed. A sample of 80
orders revealed that 70 were delivered within the promised time. At
the 0.01 significance level, can we conclude that less than 92% of
the orders are delivered in less than 10 minutes?
A.) What is the decision rule? (Negative amount should
be indicated by a minus sign. Round your answer to 2 decimal
places.)
B.) Compute...

A cell phone company offers two plans to its subscribers. At the
time new subscribers sign up, they are asked to provide some
demographic information. The mean yearly income for a sample of 38
subscribers to Plan A is $58,500 with a standard deviation of
$8,000. This distribution is positively skewed; the coefficient of
skewness is not larger. For a sample of 42 subscribers to Plan B,
the mean income is $59,100 with a standard deviation of $9,500. At
the...

Chicken Delight claims that 64% of its orders are delivered
within 10 minutes of the time the order is placed. A sample of 100
orders revealed that 58 were delivered within the promised time. At
the 0.10 significance level, can we conclude that less than 64% of
the orders are delivered in less than 10 minutes?
a. What is your decision regarding the below
hypothesis? (Round the final answer to 2 decimal
places.)
H0 is rejected if z
< ...

A cell phone company offers two plans to its subscribers. At the
time new subscribers sign up, they are asked to provide some
demographic information. The mean yearly income for a sample of 42
subscribers to Plan A is $55,500 with a standard deviation of
$8,500. This distribution is positively skewed; the coefficient of
skewness is not larger. For a sample of 40 subscribers to Plan B,
the mean income is $56,800 with a standard deviation of $8,700.
At the...

A national grocer’s magazine reports the typical shopper spends
7 minutes in line waiting to check out. A sample of 17 shoppers at
the local Farmer Jack’s showed a mean of 6.4 minutes with a
standard deviation of 4.3 minutes.
Is the waiting time at the local Farmer Jack’s less than that
reported in the national magazine? Use the 0.050 significance
level.
What is the decision rule? (Negative amount should be
indicated by a minus sign. Round your answer to...

A) McGilla Golf has decided to sell a new line of golf clubs.
The length of this project is seven years. The company has spent
$188569 on research and development for the new clubs. The plant
and equipment required will cost $2838154 and will be depreciated
on a straight-line basis. The new clubs will also require an
increase in net working capital of $125395 that will be returned at
the end of the project. The OCF of the project will...

ADVERTISEMENT

Get Answers For Free

Most questions answered within 1 hours.

ADVERTISEMENT

asked 2 minutes ago

asked 5 minutes ago

asked 7 minutes ago

asked 15 minutes ago

asked 17 minutes ago

asked 18 minutes ago

asked 20 minutes ago

asked 28 minutes ago

asked 33 minutes ago

asked 33 minutes ago

asked 40 minutes ago

asked 40 minutes ago