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Readme [11.4K]
3 years ago
8

Leila started a savings account to buy a new bike. She started with some money her grandmom gave her and added the same amount e

ach week from her babysitting earnings. What is the dependent variable?
A-the number of weeks in which she added to her savings account

B-the amount in the account when she started saving

C-the total amount she earned babysitting

D-the amount in the savings account when she finishes saving
Mathematics
1 answer:
zalisa [80]3 years ago
7 0
The answer to this question is:
<span>Leila started a savings account to buy a new bike. She started with some money her grandmom gave her and added the same amount each week from her babysitting earnings. What is the dependent variable?
</span>D-the amount in the savings account when she finishes saving
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2. A marketing firm is trying to estimate the proportion of potential car buyers that would consider
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Answer:

a. The number of people that should be in the pilot study are 600 people

b. The point estimate is 0.62\overline 6

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The margin of error for the confidence interval, E = 4% = 0.04

The confidence level = 95%

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n = \dfrac{Z^2 \times P \times (1 - P)}{E^2}

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n = \dfrac{1.96^2 \times 0.5 \times (1 - 0.5)}{0.04^2} = 600.25

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b. The point estimate for the population proportion is the sample proportion  given as follows;

\hat p = \dfrac{x}{n}

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x = The number of the statistic in the sample

n = The sample size

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The number of potential car buyers, n = 600

The number of respondent in the sample that indicated that they would consider purchasing a hybrid, x = 376

Therefore, the point estimate, for the proportion of potential car buyers that would consider buying a hybrid vehicle, \hat p = 376/600 = 0.62\overline 6

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CI=0.62 \overline 6\pm 1.96\times \sqrt{\dfrac{\hat{0.62 \overline 6}\cdot (1-\hat{0.62 \overline 6})}{600}}

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The 95% confidence interval for the true population proportion of potential buyers of hybrid vehicle, C.I. =  (0.588, 0.6654)

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MOE_\gamma = z_\gamma  \times \sqrt{\dfrac{\sigma ^2}{n} }

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MOE_\gamma = Margin of error at a given level of confidence

z_\gamma = z-score

σ = The standard deviation

n = The sample size

Therefore, the margin error can be reduced by the following two ways;

1) Reducing the confidence interval and therefore, the z-score

2) Increasing the sample size

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