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lbvjy [14]
3 years ago
15

The following time series shows the sales of a clothing store over a 10-week period. Compute the mean square error (MSE) for a 4

-week moving average forecast for the time series. Week Sales($1,000s) 1 15 2 16 3 19 4 18 5 19 6 20 7 19 8 22 9 15 10 21 Group of answer choices 21.50 7.67 15.00 16.90 2.33
Mathematics
2 answers:
Alenkasestr [34]3 years ago
8 0

Answer:

7.67

Step-by-step explanation:

Given the data :

Week_sales_Forecast ___|error|_ |error|²

1 _____15

2 ____ 16

3 ____ 19

4 ____ 18

5 ____ 19 ____ 17 _______ 2____4

6 ____ 20 ___ 18 ________2____4

7 ____ 19 ____19 ________0____0

8 ___ 22 ____ 19 ________3____9

9 ___ 15 ____ 20 ________5___ 25

10 __ 21 _____19 ________ 2 ___ 4

Mean squared error :

(4 + 4 + 0 + 9 + 25 + 4) / 6

= 46 / 6

= 7.666

= 7.67

Whitepunk [10]3 years ago
8 0

Answer:

7.67

Step-by-step explanation:

Given the data :

Week_sales_Forecast ___|error|_ |error|²

1 _____15

2 ____ 16

3 ____ 19

4 ____ 18

5 ____ 19 ____ 17 _______ 2____4

6 ____ 20 ___ 18 ________2____4

7 ____ 19 ____19 ________0____0

8 ___ 22 ____ 19 ________3____9

9 ___ 15 ____ 20 ________5___ 25

10 __ 21 _____19 ________ 2 ___ 4

Mean squared error :

(4 + 4 + 0 + 9 + 25 + 4) / 6

= 46 / 6

= 7.666

= 7.67

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Vesnalui [34]

Answer:

t=\frac{21.5-22}{\frac{1.5}{\sqrt{40}}}=-2.108    

p_v =P(t_{39}  

Conclusion  

If we compare the p value and the significance level given \alpha=0.01 we see that p_v>\alpha so we can conclude that we have enough evidence to FAIL to reject the null hypothesis, so we can conclude that the mean temperature actually its NOT significant less then 22 at 1% of signficance.  

(D) P-val=0.021, fail to reject the null hypothesis

Step-by-step explanation:

1) Data given and notation  

\bar X=21.5 represent the mean for the temperatures

s=1.5 represent the sample standard deviation

n=40 sample size  

\mu_o =22 represent the value that we want to test

\alpha=0.01 represent the significance level for the hypothesis test.  

t would represent the statistic (variable of interest)  

p_v represent the p value for the test (variable of interest)  

State the null and alternative hypotheses.  

We need to conduct a hypothesis in order to check if the mean is less than 22C, the system of hypothesis would be:  

Null hypothesis:\mu \geq 22  

Alternative hypothesis:\mu < 22  

If we analyze the size for the sample is > 30 and we don't know the population deviation so is better apply a t test to compare the actual mean to the reference value, and the statistic is given by:  

t=\frac{\bar X-\mu_o}{\frac{s}{\sqrt{n}}}  (1)  

t-test: "Is used to compare group means. Is one of the most common tests and is used to determine if the mean is (higher, less or not equal) to an specified value".  

Calculate the statistic

We can replace in formula (1) the info given like this:  

t=\frac{21.5-22}{\frac{1.5}{\sqrt{40}}}=-2.108    

P-value

We can calculate the degrees of freedom like this:

df=n-1=40-1=39

Since is a one left tailed test the p value would be:  

p_v =P(t_{39}  

Conclusion  

If we compare the p value and the significance level given \alpha=0.01 we see that p_v>\alpha so we can conclude that we have enough evidence to FAIL to reject the null hypothesis, so we can conclude that the mean temperature actually its NOT significant less then 22 at 1% of signficance.  

The best option would be:

(D) P-val=0.021, fail to reject the null hypothesis

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