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2 years ago

If p=100, what is the value of 7xp÷5

Mathematics

1 answer:

Bumek [7]2 years ago

8 0

That will be 700x
5
Now all you need to do is divide, which will then equal to 140.
C= 140

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Two catalysts may be used in a batch chemical process. Twelve batches were prepared using catalyst 1, resulting in an average yi

aalyn [17]

Answer:

a) $t=\frac{(91-85)-(0)}{2.490\sqrt{\frac{1}{12}}+\frac{1}{15}}=6.222$

$df=12+15-2=25$

$p_v =P(t_{25}>6.222) =8.26x10^{-7}$

So with the p value obtained and using the significance level given $\alpha=0.01$ we have $p_v$ so we can conclude that we have enough evidence to reject the null hypothesis.

b) $(91-85) -2.79 *2.490 \sqrt{\frac{1}{12} +\frac{1}{15}} =3.309$

$(91-85) +2.79 *2.490 \sqrt{\frac{1}{12} +\frac{1}{15}} =8.691$

Step-by-step explanation:

Notation and hypothesis

When we have two independent samples from two normal distributions with equal variances we are assuming that

$\sigma^2_1 =\sigma^2_2 =\sigma^2$

And the statistic is given by this formula:

$t=\frac{(\bar X_1 -\bar X_2)-(\mu_{1}-\mu_2)}{S_p\sqrt{\frac{1}{n_1}}+\frac{1}{n_2}}$

Where t follows a t distribution with $n_1+n_2 -2$ degrees of freedom and the pooled variance $S^2_p$ is given by this formula:

$\S^2_p =\frac{(n_1-1)S^2_1 +(n_2 -1)S^2_2}{n_1 +n_2 -2}$

This last one is an unbiased estimator of the common variance $\sigma^2$

Part a

The system of hypothesis on this case are:

Null hypothesis: $\mu_2 \leq \mu_1$

Alternative hypothesis: $\mu_2 > \mu_1$

Or equivalently:

Null hypothesis: $\mu_2 - \mu_1 \leq 0$

Alternative hypothesis: $\mu_2 -\mu_1 > 0$

Our notation on this case :

$n_1 =12$ represent the sample size for group 1

$n_2 =15$ represent the sample size for group 2

$\bar X_1 =85$ represent the sample mean for the group 1

$\bar X_2 =91$ represent the sample mean for the group 2

$s_1=3$ represent the sample standard deviation for group 1

$s_2=2$ represent the sample standard deviation for group 2

First we can begin finding the pooled variance:

$\S^2_p =\frac{(12-1)(3)^2 +(15 -1)(2)^2}{12 +15 -2}=6.2$

And the deviation would be just the square root of the variance:

$S_p=2.490$

Calculate the statistic

And now we can calculate the statistic:

$t=\frac{(91-85)-(0)}{2.490\sqrt{\frac{1}{12}}+\frac{1}{15}}=6.222$

Now we can calculate the degrees of freedom given by:

$df=12+15-2=25$

Calculate the p value

And now we can calculate the p value using the altenative hypothesis:

$p_v =P(t_{25}>6.222) =8.26x10^{-7}$

Conclusion

So with the p value obtained and using the significance level given $\alpha=0.01$ we have $p_v$ so we can conclude that we have enough evidence to reject the null hypothesis.

Part b

For this case the confidence interval is given by:

$(\bar X_1 -\bar X_2) \pm t_{\alpha/2} S_p \sqrt{\frac{1}{n_1} +\frac{1}{n_2}}$

For the 99% of confidence we have $\alpha=1-0.99 = 0.01$ and $\alpha/2 =0.005$ and the critical value with 25 degrees of freedom on the t distribution is $t_{\alpha/2}= 2.79$

And replacing we got:

$(91-85) -2.79 *2.490 \sqrt{\frac{1}{12} +\frac{1}{15}} =3.309$

$(91-85) +2.79 *2.490 \sqrt{\frac{1}{12} +\frac{1}{15}} =8.691$

7 0

2 years ago

1/8+1/3 what is the answer?

kow [346]

Answer:

11/24 as a fraction but a decimal 0.4583 ( the three is recurring)

8 0

2 years ago

Read 2 more answers

What is the greatest factor of 49 that is a prime number

Sunny_sXe [5.5K]

7.

7 squared gives 49, and it is a prime number.

5 0

2 years ago

Read 2 more answers

Question in photo, Please Help

quester [9]

The answer would be 28 feet, to get area you do length times width, 7 x 4 = 28

7 0

2 years ago

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Assume that women's heights are normally distributed with a mean given by mu equals 62.5 in,and a standard deviation given by

Misha Larkins [42]

Answer:

(a) 0.5899

(b) 0.9166

Step-by-step explanation:

Let X be the random variable that represents the height of a woman. Then, X is normally distributed with

$\mu$ = 62.5 in

$\sigma$ = 2.2 in

the normal probability density function is given by

$f(x) = \frac{1}{\sqrt{2\pi}2.2}\exp{-\frac{(x-62.5)^{2}}{2(2.2)^{2}}}$ , then

(a) $P(X < 63) = \int\limits_{-\infty}^{63}f(x) dx$ = 0.5899

(in the R statistical programming language) pnorm(63, mean = 62.5, sd = 2.2)

(b) We are seeking $P(\bar{X} < 63)$ where n = 37. $\bar{X}$ is normally distributed with mean 62.5 in and standard deviation $2.2/\sqrt{37}$ . So, the probability density function is given by

$g(x) = \frac{1}{\sqrt{2\pi}\frac{2.2}{\sqrt{37}}}\exp{-\frac{(x-62.5)^{2}}{2(2.2/\sqrt{37})^{2}}}$ , and

$P(\bar{X} < 63) = \int\limits_{-\infty}^{63}g(x)dx$ = 0.9166

(in the R statistical programming language) pnorm(63, mean = 62.5, sd = 2.2/sqrt(37))

You can use a table from a book to find the probabilities or a programming language like the R statistical programming language.

4 0

2 years ago