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Nookie1986 [14]
2 years ago
6

The following chart shows a store?s records of coat sales over two years. 2 circle graphs. A circle graph titled 2006. Top coats

is 297, parkas is 210, jackets is 213, raincoats is 137, trench coats is 103. A circle graph titled 2007. Topcoats is 223, parkas is 210, jackets is 285, raincoats is 259, trench coats is 127. If the trend shown in these graphs stays constant, what percent of the market will parkas occupy in 2008? (Round your answer to the nearest tenth.) a. 16.5% b. 19.0% c. 21.9% d. 35.7%
ANSWER IS A
Mathematics
2 answers:
Mkey [24]2 years ago
7 0

Answer:

<u>The correct answer is A. 16.5%</u>

Step-by-step explanation:

1. Let's review the information given to us to answer the question correctly and to calculate the trend:

                   2006      2007  Trend

Top Coats    297        223       -24.92%

Parkas          210        210         + 0%  

Jackets         213        285        +33.80%

Raincoats      137        259        +89.05

Trench coats 103       127         +23.30%

Total               960     1,104        +15%

2. If the trend shown in these graphs stays constant, what percent of the market will parkas occupy in 2008?

Let's calculate the percent of the market occupied by parkas.

In 2006 = 210/960 = 21.88%

In 2007 = 210/1,104 = 19.02%

In 2008 = 210/1,270 = 16.54% (210 + 0 = 210; 1,104 + 15% = 1,270)

<u>The correct answer is A. 16.5%</u>

Stella [2.4K]2 years ago
3 0

Answer:

Step-by-step explanation:

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Of the 500 sample households in the previous exercise, 7 had three or more large-screen TVs. (a) The percentage of households in
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Answer:

a) The percentage of households in the town with three or more largescreen TVs is estimated as :

The best estimation for the population proportion is :

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And that represent the 1.4%.

b) And the 95% confidence interval would be given (0.00370;0.0243).

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Step-by-step explanation:

Data given and notation  

n=500 represent the random sample taken    

X=7 represent the households with three or more large-screen TVs

\hat p=\frac{7}{500}=0.014 estimated proportion of households with three or more large-screen TVs

\alpha=0.05 represent the significance level (no given, but is assumed)    

z would represent the statistic (variable of interest)    

p= population proportion of households with three or more large-screen TVs

Part a

The percentage of households in the town with three or more largescreen TVs is estimated as :

The best estimation for the population proportion is :

\hat p=\frac{7}{500}=0.014

And that represent the 1.4%.

Part b

Yes is possible. We hav that np>10 and n(1-p)>10 so we have the assumption of normality to find the interval.

The confidence interval would be given by this formula

\hat p \pm z_{\alpha/2} \sqrt{\frac{\hat p(1-\hat p)}{n}}

For the 95% confidence interval the value of \alpha=1-0.95=0.05 and \alpha/2=0.025, with that value we can find the quantile required for the interval in the normal standard distribution.

z_{\alpha/2}=1.96

And replacing into the confidence interval formula we got:

0.014 - 1.96 \sqrt{\frac{0.014(1-0.014)}{500}}=0.00370

0.014 + 1.96 \sqrt{\frac{0.014(1-0.014)}{500}}=0.0243

And the 95% confidence interval would be given (0.00370;0.0243).

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