Jacob Lee is a frequent traveler between Los Angeles and San Diego. For the past month, he wrote down the flight times in minutes on three different airlines. The results are:

Goust Jet Red Cloudtran
51 50 52
51 53 55
52 52 60
42 62 64
51 53 61
57 49 49
47 50 49
47 49
50 58
60 54
54 51
49 49
48 49
48 50

Required:
a. Use the α =0.05 significance level and the six-step hypothesis-testing process to check if there is a difference in the mean flight times among the three airlines.
b. Develop a 95% confidence interval for the difference in the means between Goust and Cloudtran

Answers

Answer 1

Solution :

We know that

[tex]$H_0: \mu_1 = \mu_2=\mu_3$[/tex]

[tex]$H_1 :$[/tex] At least one mean is different form the others (claim)

We need to find the critical values.

We know k = 3 , N = 35, α = 0.05

d.f.N = k - 1

       = 3 - 1 = 2

d.f.D = N - k

        = 35 - 3 = 32

SO the critical value is 3.295

The mean and the variance of each sample :

Goust                      Jet red                 Cloudtran

[tex]$\overline X_1 =50.5$[/tex]           [tex]$\overline X_2 =50.07143$[/tex]        [tex]$\overline X_3 =55.71429$[/tex]

[tex]$s_1^2=19.96154$[/tex]      [tex]$s_2^2=14.68681$[/tex]         [tex]$s_3^2=36.57143$[/tex]

The grand mean or the overall mean is(GM) :

[tex]$\overline X_{GM}=\frac{\sum \overline X}{N}$[/tex]

         [tex]$=\frac{51+51+...+49+49}{35}$[/tex]

        = 52.1714

The variance between the groups

[tex]$s_B^2=\frac{\sum n_i\left( \overline X_i - \overline X_{GM}\right)^2}{k-1}$[/tex]

     [tex]$=\frac{\left[14(50.5-52.1714)^2+14(52.07143-52.1714)^2+7(55.71426-52.1714)^2\right]}{3-1}$[/tex]

   [tex]$=\frac{127.1143}{2}$[/tex]

   = 63.55714

The Variance within the groups

[tex]$s_W^2=\frac{\sum(n_i-1)s_i^2}{\sum(n_i-1)}$[/tex]

    [tex]$=\frac{(14-1)19.96154+(14-1)14.68681+(7-1)36.57143}{(14-1)+(14-1)+(7-1)}$[/tex]

   [tex]$=\frac{669.8571}{32}$[/tex]

  = 20.93304

The F-test  statistics value is :

[tex]$F=\frac{s_B^2}{s_W^2}$[/tex]

  [tex]$=\frac{63.55714}{20.93304}$[/tex]

  = 3.036212

Now since the 3.036 < 3.295, we do not reject the null hypothesis.

So there is no sufficient evidence to support the claim that there is a difference among the means.

The ANOVA table is :

Source       Sum of squares    d.f    Mean square    F

Between    127.1143                 2      63.55714          3.036212

Within        669.8571             32      20.93304

Total           796.9714            34


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PLEASE HELP PLEASE

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Answers

Answer:

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Step-by-step explanation:

Answer:   18 different outfit combinations

=============================================================

Explanation:

There are 3 colors to pick for the sweaters (red, green, purple).There are 3 colors for the skirts (tan, white, black)There are 2 colors for the shoes (gold, black)

Multiply out those values mentioned to get the final answer: 3*3*2 = 9*2 = 18.

There are 18 different outfit combos.

---------------

Further explanation (optional section):

Let's say we only worried about the sweaters and shoes, and for now there's only one color for the skirt. We have 3 colors for the sweaters and 2 for the shoes. Imagine a table with 3 rows and 2 columns. Each row is a different color for the sweater and each column is a different color for the shoe. There are 3*2 = 6 cells inside the table. Each cell is a different outfit.

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Answer: ...BEDMAS..

steps,..

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i hope this helps!! ^_^

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Answers

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When the numerator exceeds or is equal to the denominator, the fraction is said to be inappropriate. 5/2 and 8/5, for instance, are incorrect fractions. The numerator and denominator are the two components of any fraction.

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[tex]3\dfrac{1}{9}[/tex]

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≈ 107.33...

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To find the % then we have too times the 16 /12% by something right?

After some guessing their is my answer

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If you are reading this hope you have a wonderful day!


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Answers

Answer:

9

Step-by-step explanation:

Subtract 5 from each side and get -9x = -81.

Divide by -9.

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Answer:

x=9

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5-9x=-76

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Answer:

About what and give me the picture

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