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3 years ago

10

Use the given data to find the minimum sample size required to estimate the population proportion. Margin of error: 0.028; confi

dence level: 99%; pp is unknown.
a. 1116
b. 2223
c. 1939
d. 2115

1 answer:

TiliK225 [7]3 years ago

3 0

The answer I think it’s number b

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HI! I HAVE A TEST ON MONDAY AND IM TRYING TO STUDY, IM KINDA STUCK ON HOW TO DO THIS PROBLEM..

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Answer: See below

Step-by-step explanation:

number of yellow pieces/number of blue pieces

= 14/24

= 7/12 --> 7 to 12 --> 7:12

The ratio that represents the number of yellow pieces to the total number of pieces is part-to-whole

number of yellow pieces/total number of pieces

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2 years ago

Every month, Mr. Leroy pays his rent through automatic payments (withdrawals) from his

GarryVolchara [31]

Answer:

-445/month

Step-by-step explanation:

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3 years ago

According to 2017 Belgium Labor data information: Adult Population in 2017= 4500, Number of employed in 2017= 2800, and Number o

maks197457 [2]

The labor-force participation rate of Belgium in 2017 was 77.78%

1.

Given:

Adult population = 4500

Number of employed in 2017 = <em>2800</em>

Number of Unemployed in 2017 = 700

Labor force participation rate = (employed + unemployed) / total population × 100

= (2800 + 700) / 4,500 × 100

= 3,500 / 4500 × 100

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= 77.77777777777777%

Approximately, 77.78%

2.

Given:

Adult Population in 2016 = 3500

Number of employed in 2016 = 1800

Number of Unemployed in 2016= 600

Unemployment rate = unemployed population/ total labor force × 100

= 600 / 3,500 × 100

= 0.171428571428571 × 100

= 17.14285714285714%

Approximately,

17.14 %

Therefore, the unemployment rate of Japan in 2016 was 17.14%

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2 years ago

Write 931 x 10-5 in standard notation.

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Answer:

0.000093

Step-by-step explanation:

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3 years ago

Read 2 more answers

In 2008, the average household debt service ratio for homeowners was 13.2. The household debt service ratio is the ratio of debt

ankoles [38]

Answer:

$t=\frac{13.88-13.2}{\frac{3.14}{\sqrt{44}}}=1.436$

$df=n-1=44-1=43$

$p_v =P(t_{(43)}>1.436)=0.079$

We see that the p value i higher than the significance level so then we FAIL to reject the null hypothesis and we can conclude that the true mean is not significantly higher than 13.2 *the value of 2008 ).

Step-by-step explanation:

Information given

$\bar X=13.88$ represent the sample mean

$s=3.14$ represent the sample standard deviation

$n=44$ sample size

$\mu_o =13.2$ represent the value that we want to test

$\alpha=0.05$ represent the significance level

t would represent the statistic (variable of interest)

$p_v$ represent the p value for the test

Hypothesis to test

We want to conduct a hypothesis in order to check if the true mean has increased from 2008 , and the system of hypothesi are:

Null hypothesis: $\mu \leq 13.2$

Alternative hypothesis: $\mu > 13.2$

The statistic for this case is:

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

Calculating the statistic

Replacing the info given we got:

$t=\frac{13.88-13.2}{\frac{3.14}{\sqrt{44}}}=1.436$

P-value

The degrees of freedom are:

$df=n-1=44-1=43$

Since is a right tailed test the p value is:

$p_v =P(t_{(43)}>1.436)=0.079$

Decision

We see that the p value i higher than the significance level so then we FAIL to reject the null hypothesis and we can conclude that the true mean is not significantly higher than 13.2 *the value of 2008 ).

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4 years ago