What was the state is the probability that a given piece of candy in another bag is yellow ?

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.

8 0

3 years ago

Two people spend $44 at a fair. There is an entrance fee of x dollars per person and each ride costs $3.50. If they went on the

almond37 [142]

I think its B
that's THINK i'm not completely sure

7 0

2 years ago

Which relation is a function? A) x = y2 - 3 B) x = y - 4x2 C) y2 = 4x - 6 D) x2 = y2 - 10

Nitella [24]

The answer is letter b

5 0

3 years ago

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The box plots summarize the data for heights of players on a national

andreev551 [17]

(Diagram of the box plots is attached below)

Options:

A. 75% of the volleyball players’ heights are equal to or greater than the median basketball players’ heights

B. 75% of the volleyball players’ heights are equal to or greater than the median basketball players’ heights

C. 25% of the basketball players’ heights are equal to or greater than the median volleyball players’ heights

D. 75% of the basketball players’ heights are equal to or greater than the median volleyball players’ heights

Answer:

D. 75% of the basketball players’ heights are equal to or greater than the median volleyball players’ heights

Step-by-step Explanation:

Based on the box plots given in the diagram attached below, the median height of volleyball players is 79.

Also, from the data set of basketball players, we can approximately infer that the heights from Q2, Q3, and Q4 make up 75% of the data set of the heights of basketball players.

Heights at Q2 in the data set of basketball players starts from 79, which is equal to the median height of volleyball players.

Therefore, from the displays, we can conclude that "75% of the basketball players’ heights are equal to or greater than the median volleyball players’ heights".

8 0

3 years ago

Oop help plzzzzz this is ALOTE OF MY GRADE! HELP RIGHT AWAY ILL GIVE 20 POINTS PLZ HELP

sertanlavr [38]

Answer:

its B ITHINK

Step-by-step explanation:

7 0

3 years ago

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