V= <img src="https://tex.z-dn.net/?f=v%3D%20%5Cpi%20r%20%5E%7B2%7D%20h%20%20for%20%20%202%20" id="TexFormula1" title="v= \pi r ^

You have to move the decimal point from the end to between the 6 and 5 then count how many spaces you moved it because this will become the power on the 10.
6.5 x 10^7

3 0

3 years ago

A researcher examines 27 water samples for iron concentration. The mean iron concentration for the sample data is 0.802 cc/cubic

Delvig [45]

Answer:

90% confidence interval for the population mean iron concentration is [0.771 , 0.832].

Step-by-step explanation:

We are given that a researcher examines 27 water samples for iron concentration. The mean iron concentration for the sample data is 0.802 cc/cubic meter with a standard deviation of 0.093.

Firstly, the pivotal quantity for 90% confidence interval for the population mean iron concentration is given by;

P.Q. = $\frac{\bar X - \mu}{\frac{s}{\sqrt{n} } }$ ~ $t_n_-_1$

where, $\bar X$ = sample mean iron concentration = 0.802 cc/cubic meter

s = sample standard deviation = 0.093

n = number of water samples = 27

$\mu$ = population mean

<em>Here for constructing 90% confidence interval we have used t statistics because we don't know about population standard deviation.</em>

So, 90% confidence interval for the population mean, $\mu$ is ;

P(-1.706 < $t_2_6$ < 1.706) = 0.90 {As the critical value of t at 26 degree of

freedom are -1.706 & 1.706 with P = 5%}

P(-1.706 < $\frac{\bar X - \mu}{\frac{s}{\sqrt{n} } }$ < 1.706) = 0.90

P( $-1.706 \times {\frac{s}{\sqrt{n} } }$ < ${\bar X - \mu}$ < $1.706 \times {\frac{s}{\sqrt{n} } }$ ) = 0.90

P( $\bar X -1.706 \times {\frac{s}{\sqrt{n} }$ < $\mu$ < $\bar X +1.706 \times {\frac{s}{\sqrt{n} }$ ) = 0.90

<u>90% confidence interval for</u> $\mu$ = [ $\bar X -1.706 \times {\frac{s}{\sqrt{n} }$ , $\bar X +1.706 \times {\frac{s}{\sqrt{n} }$ ]

= [ $0.802 -1.706 \times {\frac{0.093}{\sqrt{27} }$ , $0.802 +1.706 \times {\frac{0.093}{\sqrt{27} }$ ]

= [0.771 , 0.832]

Therefore, 90% confidence interval for the population mean iron concentration is [0.771 , 0.832].

4 0

3 years ago

A researcher reports survey results by stating that the standard error of the mean is 25 the population standard deviation is 40

bezimeni [28]

Answer:

a) A sample of 256 was used in this survey.

b) 45.14% probability that the point estimate was within ±15 of the population mean

Step-by-step explanation:

This question is solved using the normal probability distribution and the central limit theorem.

Normal probability distribution

When the distribution is normal, we use the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$ , the zscore of a measure X is given by:

$Z = \frac{X - \mu}{\sigma}$

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

Central Limit Theorem

The Central Limit Theorem estabilishes that, for a normally distributed random variable X, with mean $\mu$ and standard deviation $\sigma$ , the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean $\mu$ and standard deviation $s = \frac{\sigma}{\sqrt{n}}$ .