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Delicious77 [7]
3 years ago
8

Please help, thank you!

Mathematics
1 answer:
Blababa [14]3 years ago
8 0

The formula for calculating simple interest is

I = PRT

I = Interest

P = Principal or the amount deposited ($600)

R = Rate of Interest (1.5%)

T = Time (6 yrs)

I = (600) (1.50) (6)

Multiply the 3 factors and you will have your answer.

good luck

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Step-by-step explanation:

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An automobile manufacturer has given its van a 59.5 miles/gallon (MPG) rating. An independent testing firm has been contracted t
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Answer:

The pvalue of the test is 0.0124 < 0.1, which means that there is sufficient evidence at the 0.1 level to support the testing firm's claim.

Step-by-step explanation:

An automobile manufacturer has given its van a 59.5 miles/gallon (MPG) rating. An independent testing firm has been contracted to test the actual MPG for this van since it is believed that the van has an incorrect manufacturer's MPG rating:

At the null hypothesis, we test if the mean is the same, that is:

H_0: \mu = 59.5

At the alternate hypothesis, we test that it is different, that is:

H_a: \mu \neq 59.5

The test statistic is:

z = \frac{X - \mu}{\frac{\sigma}{\sqrt{n}}}

In which X is the sample mean, \mu is the value tested at the null hypothesis, \sigma is the standard deviation and n is the size of the sample.

59.5 is tested at the null hypothesis:

This means that \mu = 59.5

After testing 250 vans, they found a mean MPG of 59.2. Assume the population standard deviation is known to be 1.9.

This means that n = 250, X = 59.2, \sigma = 1.9

Value of the test statistic:

z = \frac{X - \mu}{\frac{\sigma}{\sqrt{n}}}

z = \frac{59.2 - 59.5}{\frac{1.9}{\sqrt{250}}}

z = -2.5

Pvalue of the test and decision:

The pvalue of the test is the probability of finding a mean that differs from 59.5 by at least 0.3, which is P(|Z|>-2.5), which is 2 multiplied by the pvalue of Z = -2.5.

Looking at the z-table, Z = -2.5 has a pvalue of 0.0062

2*0.0062 = 0.0124

The pvalue of the test is 0.0124 < 0.1, which means that there is sufficient evidence at the 0.1 level to support the testing firm's claim.

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Soft drink consumption in New Zealand. A survey commissioned by the Southern Cross Healthcare Group reported that 16% of New Zea
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<h2>Answer with explanation:</h2>

Given : The proportion of New Zealanders consume five or more servings of soft drinks per week :\hat{p}= 0.16

a) The number of survey respondents reported that they consume five or more servings of soft drinks per week = 2006

b) Confidence interval for population proportion (p) :

\hat{p}\pm z^*\sqrt{\dfrac{\hat{p}(1-\hat{p})}{n}}

, where \hat{p} = Sample proportion.

n= Sample size.

z* = Critical value.

For n= 2006 , \hat{p}= 0.16 and critical value for 95% confidence interval : z* = 1.96

Then , the required confidence interval will be :

0.16\pm (1.96)\sqrt{\dfrac{ 0.16(1- 0.16)}{2006}}

0.16\pm (1.96)\sqrt{ 0.000066999002991}

0.16\pm (1.96)(0.00818529186963)

0.16\pm 0.0160431720645

(0.16-0.0160431720645,\ 0.16+0.0160431720645)

(0.143956827935,\ 0.176043172064)\approx(0.144,\ 0.176)

i.e. A 95% confidence interval for the proportion of New Zealanders who report that they consume five or more servings of soft drinks per week. = (0.144,\ 0.176)

c) (0.144,\ 0.176)=(0.144\times100\%,\ 0.176\times100\%)

=(14.4\%,\ 17.6\%)176\times100\%)[/tex]

d) The estimate might be biased because the survey is taken online that means offline New Zealanders are out of consideration.

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