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2 years ago

Subtract integers 9 - (10)

Mathematics

1 answer:

BartSMP [9]2 years ago

4 0

Step-by-step explanation:

9-(10) = 9-10 = -1

Hope this helps

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Find the size of each of two samples (assume that they are of equal size) needed to estimate the difference between the proporti

hodyreva [135]

Answer:

The sample size n = 4225

Step-by-step explanation:

We will use maximum error formula = $\frac{z_{a} S.D}{\sqrt{n} }$

but we will find sample size "n"

$\sqrt{n} = \frac{z_{a} S.D}{maximum error}$

Squaring on both sides , we get

$n = (\frac{z_{a} S.D}{maximum error})^2$

Given 99% confidence interval (z value) = 2.56

given maximum error = 0.02

$n = (\frac{z_{a} p(1-p)}{maximum error})^2$

n≤ $(\frac{2.56X\frac{1}{2} }{0.02}) ^{2}$ ( here S.D = p(1-p) ≤ 1/2

on simplification , we get n = 4225

<u>Conclusion</u>:

The sample size of two samples is n = 4225

<u>verification</u>:-

We will use maximum error formula = $\frac{z_{a} S.D}{\sqrt{n} }$ = $\frac{2.56 1/2}{\ \sqrt{4225} }$ = 0.0196

substitute all values and simplify we get maximum error is 0.02

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2 years ago

How do u do order pairs

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2 years ago

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An article describes an experiment to determine the effectiveness of mushroom compost in removing petroleum contaminants from so

alexgriva [62]

Solution :

Let $p_1$ and $p_2$ represents the proportions of the seeds which germinate among the seeds planted in the soil containing $3\%$ and $5\%$ mushroom compost by weight respectively.

To test the null hypothesis $H_0: p_1=p_2$ against the alternate hypothesis $H_1:p_1 \neq p_2$ .

Let $\hat p_1, \hat p_2$ denotes the respective sample proportions and the $n_1, n_2$ represents the sample size respectively.

$$\hat p_1 = \frac{74}{155} = 0.477419$

$n_1=155$

$$p_2=\frac{86}{155}=0.554839$

$n_2=155$

The test statistic can be written as :

$$z=\frac{(\hat p_1 - \hat p_2)}{\sqrt{\frac{\hat p_1 \times (1-\hat p_1)}{n_1}} + \frac{\hat p_2 \times (1-\hat p_2)}{n_2}}}$

which under $H_0$ follows the standard normal distribution.

We reject $H_0$ at $0.05$ level of significance, if the P-value or if $|z_{obs}|>Z_{0.025}$

Now, the value of the test statistics = -1.368928

The critical value = $\pm 1.959964$

P-value = $P(|z|> z_{obs})= 2 \times P(z< -1.367928)$

$=2 \times 0.085667$

= 0.171335

Since the p-value > 0.05 and $|z_{obs}| \ngtr z_{critical} = 1.959964$ , so we fail to reject $H_0$ at $0.05$ level of significance.

Hence we conclude that the two population proportion are not significantly different.

Conclusion :

There is not sufficient evidence to conclude that the $\text{proportion}$ of the seeds that $\text{germinate differs}$ with the percent of the $\text{mushroom compost}$ in the soil.