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soldier1979 [14.2K]
3 years ago
7

A survey of 35 people was conducted to compare their self-reported height to their actual height. The difference between reporte

d height and actual height was calculated. You're testing the claim that the mean difference is greater than 0.7. From the sample, the mean difference was 0.95, with a standard deviation of 0.44. Calculate the test statistic, rounded to two decimal place
Mathematics
1 answer:
Over [174]3 years ago
3 0

Answer:

The test statistic is t = 3.36.

Step-by-step explanation:

You're testing the claim that the mean difference is greater than 0.7.

At the null hypothesis, we test if the mean difference is of 0.7 or less, that is:

H_0: \mu \leq 0.7

At the alternate hypothesis, we test if the mean difference is greater than 0.7, that is:

H_1: \mu > 0.7

The test statistic is:

t = \frac{X - \mu}{\frac{s}{\sqrt{n}}}

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

0.7 is tested at the null hypothesis:

This means that \mu = 0.7

A survey of 35 people was conducted to compare their self-reported height to their actual height.

This means that n = 35

From the sample, the mean difference was 0.95, with a standard deviation of 0.44.

This means that X = 0.95, s = 0.44

Calculate the test statistic

t = \frac{X - \mu}{\frac{s}{\sqrt{n}}}

t = \frac{0.95 - 0.7}{\frac{0.44}{\sqrt{35}}}

t = 3.36

The test statistic is t = 3.36.

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  b = 70000 +10000 = 80,000

The amounts invested in stocks, bonds, and CDs were $45,000, $80,000, and $10,000, respectively.

_____

Alternatively, you can reduce the augmented matrix for this problem to row-echelon form using any of several calculators or on-line sites. That matrix is ...

\left[\begin{array}{ccc|c}1&1&1&135000\\0.0275&0.045&0.104&8555\\-1&1&0&70000\end{array}\right]

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