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3 years ago

How to find two fractions between 1/2 and 1/3

1 answer:

4 0

1. write 1/2 and 1/3 as fractions with a common denominator.

the common denominator must be a multiple of both 2 and 3, for example 6, 12, 24 etc...

let the common denominator be 24

(1/2)(12/12)=12/24

(1/3)(8/8)=8/24

2. so , 9/24, 10/24, 11/24 are all fractions between 1/2 and 1/3

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a) $\bar d = \frac{12.6 +14.4 +14.7 +14.5 +15.2 +13.5}{6}=14.15$

b) $ME=2.57 \frac{0.940}{\sqrt{6}}=0.986$

c) $14.15 - 2.57 \frac{0.940}{\sqrt{5}}=13.164$

$14.15 + 2.57 \frac{0.940}{\sqrt{5}}=15.136$

The 95% confidence interval is given by (13.164.15.136)

Step-by-step explanation:

Previous concepts

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".

The margin of error is the range of values below and above the sample statistic in a confidence interval.

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

Part a

12.6 14.4 14.7 14.5 15.2 13.5.

Assuming that the men in this study are representative of the population of all men, what is an estimate of the population mean increase in height after three full days in bed?

For this case the best estimate for the mean is the average given by this formula:

$\bar X = \frac{\sum_{i=1}^n X_i}{n}$

For our case we are taking differences so would be the mean of differences and we got:

$\bar d = \frac{12.6 +14.4 +14.7 +14.5 +15.2 +13.5}{6}=14.15$

Part b

Assuming 95 % of confidence level. In order to find the critical value is important to mention that we don't know about the population standard deviation, so on this case we need to use the t distribution. Since our interval is at 95% of confidence, our significance level would be given by $\alpha=1-0.95=0.05$ and $\alpha/2 =0.025$ . The degrees of freedom are given by: