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3 years ago

In a test of physical fitness, a group of men ages 65 and older from a local

Mathematics

2 answers:

olga nikolaevna [1]3 years ago

7 0

Answer: x = 24.5

Step-by-step explanation: Sample mean is the average value of a fraction of the whole population. It is also an estimated of how or what the whole population is doing. To determine it, there is the following formula:

x = ∑ $\frac{x_{i} }{N}$ , where:

x is sample mean;

$x_{i}$ is each of the values;

N is the quantity the sample has;

From the given data, N = 16, so,

x = $\frac{24+25+10+23+18+29+47+32+19+24+20+28=21+20+27+25}{16}$

x = 24.5

The sample mean is x = 24.5

seraphim [82]3 years ago

6 0

Answer:

The sample mean = 24.5

Step-by-step explanation:

Solution:

Using the one sample z test

Since the test is a two tailed test

Null hypothesis, H₀: μ = 20.

Alternative hypothesis, Ha: µ ≠ 20

The test statistic formula is given by:

$z = \frac{\bar{x} - \mu}{\frac{\sigma}{\sqrt{n} } }$

The population mean, µ = 20

The sample mean, $\bar{x} = \frac{\sum x}{n}$

n = 16

$\bar {x} = \frac{24 + 25 + 10 + 23 + 18 + 29 + 47 + 32 + 19+24+20+28+21+20+27+25}{16}$

$\bar {x} = \frac{392}{16} \\\bar {x} = 24.5$

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A public health official is planning for the supplyof influenza vaccine needed for the upcoming flu season. She wants to estimat

Marizza181 [45]

Answer:

$n=\frac{0.5(1-0.5)}{(\frac{0.04}{1.64})^2}=420.25$

And rounded up we have that n=421

Step-by-step explanation:

We know that the sample proportion have the following distribution:

$\hat p \sim N(p,\sqrt{\frac{p(1-p)}{n}})$

In order to find the critical value we need to take in count that we are finding the interval for a proportion, so on this case we need to use the z distribution. Since our interval is at 90% of confidence, our significance level would be given by $\alpha=1-0.90=0.1$ and $\alpha/2 =0.05$ . And the critical value would be given by:

$z_{\alpha/2}=-1.64, z_{1-\alpha/2}=1.64$

The margin of error for the proportion interval is given by this formula:

$ME=z_{\alpha/2}\sqrt{\frac{\hat p (1-\hat p)}{n}}$ (a)

And on this case we have that $ME =\pm 0.04$ and we are interested in order to find the value of n, if we solve n from equation (a) we got:

$n=\frac{\hat p (1-\hat p)}{(\frac{ME}{z})^2}$ (b)

We assume that a prior estimation for p would be $\hat p =0.5$ since we don't have any other info provided. And replacing into equation (b) the values from part a we got: