1answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
iris [78.8K]
3 years ago
10

Many airline companies have begun implementing fees for checked bags. Economic theory predicts that passengers will respond to t

he increase in the price of a checked bag by substituting carry-on bags for checked bags. As a result, the mean weight of a passenger's carry-on items is expected to increase after the implementation of the checked-bag fee.
Suppose that a particular airline's passengers had a mean weight for their carry-on items of 16 pounds, the FAA standard average weight, before the implementation of the checked-bag fee.

The airline conducts a hypothesis test to determine whether the current mean weight of its passengers' carry-on items is more than 16 pounds. It selects a random sample of 67 passengers and weighs their carry-on items. The sample mean is x = 17.1 pounds, and the sample standard deviation is
s = 6.0pounds. The airline uses a significance level of α =.05 to conduct its hypothesis test.

a. The hypothesis test is _____ test.

- The test statistic follows a _____ distribution. The value of the test statistic is_____.

b. Use the relevant statistical Distributions table to develop the critical value rejection rule.

According to the critical value approach, the rejection rule is:

A. Reject H0 if t ≤ − 1.997 or t ≥ 1.997
B. Reject H0 if z ≥ 1.645
C. Reject H0 if t ≥ 1.668
D. Reject H0 if t ≤ −1.668

c. The p-value is_____.

d. Using the critical value approach, the null hypothesis is _____, because _____ Using the p-value approach, the null hypothesis is_____, because_____ Therefore, you _____ conclude that the mean weight of the airline's passengers' carry-on items has increased after the implementation of the checked-bag fee.
Mathematics
1 answer:
valentinak56 [21]3 years ago
4 0

Answer:

Step-by-step explanation:

a. The hypothesis test is  one tailed_____ test.

(Because we check whether sample weight is greater than hence one tailed or right tailed)

The test statistic follows a __t___ distribution.(Because only sample std deviation s is known)

The value of the test statistic is___Mean difference/Std error =\frac{17.1-16}{\frac{6}{\sqrt{67} } } \\=1.58__

b. df = 66

Reject H0 if t ≥ 1.668

c. The p-value is_____0.059444

d. Using the critical value approach, the null hypothesis is _accepted____, because __t <1.668___ Using the p-value approach, the null hypothesis is__accepted___, because__p value <0.05 our significance level.___ Therefore, you __may___ conclude that the mean weight of the airline's passengers' carry-on items has increased after the implementation of the checked-bag fee.

You might be interested in
What is the volume of this figure?<br><br> Enter your answer in the box.<br><br> m³
IRINA_888 [86]

Answer:

Im pretty sure the answer is 497; i did the math :)

Step-by-step explanation:

7 0
2 years ago
The diameter of a circle is 18 kilometers. What is the circle's circumference? d=18 km Use 3.14 for n. kilometers​
disa [49]
Yeah I got you bro I got a friend that knows his stuff but by the time in the tsunami
8 0
3 years ago
2 _ 5 _ 3 = 17 <br> Fill in the blank
siniylev [52]

Answer:

2+5*3=17

Step-by-step explanation:

The blanks should be filled in with a plus (+) and a times sign (*), respectively. To solve this question you must understand PEMDAS. This stands for paratheses, exponents, multiplication, division, addition, subtraction. When solving an equation, operations should be done in this order. Therefore, numbers should be multiplied before they are added.

So, to solve this equation multiply 5*3; this equals 15. Then add 2, which equals 17. And 17=17, so the equation must be true.

6 0
3 years ago
Read 2 more answers
textRequest reports that adults 18–24 years old send and receive 128 texts every day. Suppose we take a sample of 25–34 year old
Butoxors [25]

Testing the hypothesis, we have that:

a)

The null hypothesis is: H_0: \mu = 128

The alternative hypothesis is: H_1: \mu \neq 128

b) The p-value of the test is of 0.1212.

c) Since the <u>p-value of the test is of 0.1212 > 0.05</u>, we cannot conclude that the mean daily number of texts for 25–34 year olds differs from the population daily mean number of texts for 18–24 year olds.

d) Since <u>|z| = 1.55 < 1.96</u>, we cannot conclude that the mean daily number of texts for 25–34 year olds differs from the population daily mean number of texts for 18–24 year olds.

Item a:

At the null hypothesis, we <u>test if the mean is the same</u>, that is, of 128 texts every day, hence:

H_0: \mu = 128

At the alternative hypothesis, we <u>test if the mean is different</u>, that is, different of 128 texts every day, hence:

H_1: \mu \neq 128

Item b:

We have the <u>standard deviation for the population</u>, thus, the z-distribution is used. The test statistic is given by:

z = \frac{\overline{x} - \mu}{\frac{\sigma}{\sqrt{n}}}

The parameters are:

\overline{x} is the sample mean.

\mu is the value tested at the null hypothesis.

\sigma is the standard deviation of the population.

n is the sample size.

For this problem, the values of the parameters are: \overline{x} = 118.6, \mu = 128, \sigma = 33.17, n = 30

Hence, the value of the test statistic is:

z = \frac{\overline{x} - \mu}{\frac{\sigma}{\sqrt{n}}}

z = \frac{118.6 - 128}{\frac{33.17}{\sqrt{30}}}

z = -1.55

Since we have a two-tailed test, as we are testing if the mean is different of a value, the p-value is P(|z| < 1.55), which is 2 multiplied by the p-value of z = -1.55.

Looking at the z-table, z = -1.55 has a p-value of 0.0606

2(0.0606) = 0.1212

The p-value of the test is of 0.1212.

Item c:

Since the <u>p-value of the test is of 0.1212 > 0.05</u>, we cannot conclude that the mean daily number of texts for 25–34 year olds differs from the population daily mean number of texts for 18–24 year olds.

Item d:

Using a z-distribution calculator, the critical value for a <u>two-tailed test</u> with <u>95% confidence level</u> is |z| = 1.96.

Since <u>|z| = 1.55 < 1.96</u>, we cannot conclude that the mean daily number of texts for 25–34 year olds differs from the population daily mean number of texts for 18–24 year olds.

A similar problem is given at brainly.com/question/25369247

4 0
3 years ago
Help please! thank you!
Andreyy89

Answer:

The probability that a point randomly chose in the rectangle is in the triangle is \frac{1}{8}, so the answer would be A

Step-by-step explanation:

To find the probability of a random point being in the triangle, we must find the area of the triangle and the area of the rectangle.

The area of the rectangle can be found by multiplying the width by the height, so the equation would be

8 *4 = 32

The area of the the triangle can be found by the formula A = \frac{1}{2} bh

This means that the equation would look like

\frac{1}{2} *2*4=4

Now that we have both the area of the triangle and the rectangle, we can create a ration of the area of the triangle to that of the rectangle. It would look like this

\frac{4}{32}

This would simplify to \frac{1}{8}

8 0
3 years ago
Read 2 more answers
Other questions:
  • F(x)=2x^2+4x+5<br> Find : f(a)=
    7·1 answer
  • How do you make an equivalent fraction to another fraction?
    15·1 answer
  • HELP 20 PPOINTS
    8·2 answers
  • The mass of an electron is
    9·1 answer
  • Which pair of ratios is equivalent
    13·1 answer
  • What is the inverse function of f(x) = -2?<br> Hurry please
    10·1 answer
  • Plzzz helps me for once!!!!!! Find the slope of the graphs below.
    13·1 answer
  • Select all ratios that are in their simplest form.<br> A 11:12 B 4:28 C 7:14 D 26:3 E 11:29
    12·1 answer
  • Find the slope and y-intercept for the graph.<br><br> God bless!
    9·1 answer
  • Sue has 48 dolls. She gives 5/8 of them to her sister. How many dolls does she give her?
    10·2 answers
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!