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KatRina [158]
3 years ago
14

The forecast for week 13 is 747. ​(Enter your response rounded to the nearest whole​ number.) Specify the mean absolute deviatio

n​ (MAD) and mean absolute percent error​ (MAPE). ​(Enter your responses rounded to two decimal​ places.)
Mathematics
1 answer:
son4ous [18]3 years ago
7 0

Answer:

a) MAD without the forecast = 43.71

MAD with the forecast = 45.59

b) MAPE = 0.097 = 9.7%

Step-by-step explanation:

642 602 656 747 663 618 731 726 679 737 664 740

Mean Absolute deviation is given as

MAD = [Σ|x - μ|]/N

We first calculate the mean

Mean = Σx/N

Mean = (642+602+656+747+663+618+731+726+679+737+664+740)/12

Mean = (8205/12) = 683.75

absolute deviations from the mean

|642-683.75| + |602-683.75| + |656-683.75| + |747-683.75| + |663-683.75| + |618-683.75| + |731-683.75| + |726-683.75| + |679-683.75| + |737-683.75| + |664-683.75| + |740-683.75|

Absolute deviations from the mean

41.75+81.75+27.75+63.25+20.75+65.75+47.25+42.25+4.75+53.25+19.75+56.25

= 524.5

MAD = [Σ|x - μ|]/N

MAD = (524.5/12)

MAD = 43.71

If we include the forecast on the 13th week

642 602 656 747 663 618 731 726 679 737 664 740 747

Mean = (8205+747)/13

Mean = 688.62

absolute deviations from the mean

|642-688.62| + |602-688.62| + |656-688.62| + |747-688.62| + |663-688.62| + |618-688.62| + |731-688.62| + |726-688.62| + |679-688.62| + |737-688.62| + |664-688.62| + |740-688.62| + |747-688.62|

Absolute deviations from the mean

= 46.62+86.62+32.62+58.38+25.62+70.62+42.38+37.38+9.62+48.38+24.62+51.38+58.38

= 592.62

MAD = (592.62/13)

MAD = 45.59.

b) MAPE is used to check forecast errors

MAPE = (1/N) Σ [|x-f|/x]

where N = Sample size = 12

x = each variable

f = the forecasted value = 747

x = 642, 602, 656, 747, 663, 618, 731, 726, 679, 737, 664, 740

Σ [|x-f|/x]

= 0.164+0.241+0.139+0+0.112+0.209+0.022+0.029+0.100+0.014+0.125+0.009

Σ [|x-f|/x] = 1.164

MAPE = (1.164/12)

MAPE = 0.097 = 9.7%

Hope this Helps!!!

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The given data are;

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Chain B 4.60, 4.65, 3.85, 4.00, 4.80, 4.00, 4.50, 3.65

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The mean hourly rate for fast-food Chain A, \overline x_1 = 4.25

The standard deviation hourly rate for fast-food Chain A, s₁ = 0.457478

The mean hourly rate for fast-food Chain B, \overline x_2 = 4.25625

The standard deviation hourly rate for fast-food Chain B, s₂ = 0.429649

The significance level, α = 5%

The null hypothesis, H₀:  \overline x_1 = \overline x_2

The alternative hypothesis, Hₐ:  \overline x_1 ≠ \overline x_2

The pooled variance, S_p^2, is given as follows;

S_p^2 = \dfrac{s_1^2 \cdot (n_1 - 1) + s_2^2\cdot (n_2-1)}{(n_1 - 1)+ (n_2 -1)}

Therefore, we have;

S_p^2 = \dfrac{0.457478^2 \cdot (8 - 1) + 0.429649^2\cdot (8-1)}{(8 - 1)+ (8 -1)} \approx 0.19682

The test statistic is given as follows;

t=\dfrac{(\bar{x}_{1}-\bar{x}_{2})}{\sqrt{S_{p}^{2} \cdot \left(\dfrac{1 }{n_{1}}+\dfrac{1}{n_{2}}\right)}}

Therefore, we have;

t=\dfrac{(4.25-4.25625)}{\sqrt{0.19682 \times \left(\dfrac{1 }{8}+\dfrac{1}{8}\right)}} \approx -0.028176

The degrees of freedom, df = n₁ + n₂ - 2 = 8 + 8 - 2 = 14

At 5% significance level, the critical t = 2.145

Therefore, given that the absolute value of the test statistic is less than the critical 't', we fail to reject the null hypothesis and it can be concluded that at 5% significance level that chain A pays the same as chain B for the job under consideration

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