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2 years ago

I require help lol so help me pleasw

Mathematics

1 answer:

r-ruslan [8.4K]2 years ago

3 0

Answer/Step-by-step explanation:

Mean is a measure of center tendency which can be said to be the average value of a set of data. It can be referred to as the commonest value in a data set. It is the sum of all data values in a given set divided by the number of values.

For example, a test score for a class of 5 is given as 5, 6, 2, 4, 7

The mean = sum of all values/number of data values

= (5 + 6 + 2 + 4 + 7)/5

= 24/5

Mean = 4.8

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The students' attitudes were measured by administering the Computer Anxiety Rating Scale (CARS). A random sample of 11 nursing s

Fofino [41]

Answer:

There is enough evidence to claim that population mean 1 is less than population 2 at the 0.05 significance level.

Step-by-step explanation:

We are given the following in the question:

Group 1:

$\mu_1 = 63.3\\\sigma_1 = 3.7\\n_1 = 11$

Group 2:

$\mu_2 = 70.2\\\sigma_2 = 6.6\\n_2 = 13$

Alpha, α = 0.05

First, we design the null and the alternate hypothesis

$H_{0}: \mu_1 \geq \mu_2\\H_A: \mu_1 < \mu_2$

Since, the population variances are equal and that the two populations are normally distributed, we use t-test(pooled test) for difference of two means.

Formula:

Pooled standard deviation

$s_p = \sqrt{\displaystyle\frac{(n_1-1)\sigma_1^2 + (n_2-1)\sigma_2^2 }{n_1 + n_2 - 2}}$

$t_{stat} = \displaystyle\frac{\mu_1-\mu_2}{s_p\sqrt{\frac{1}{n_1}+\frac{1}{n_2}}}$

$\text{Degree of freedom} = n_1 + n_2 - 2$

Putting all the values we get:

$s_p = \sqrt{\displaystyle\frac{(11-1)(3.7)^2 + (13-1)(6.6)^2 }{11 + 13 - 2}} = \sqrt{29.9827} = 5.48$

$t_{stat} = \displaystyle\frac{63.3-70.2}{5.48\sqrt{\frac{1}{11}+\frac{1}{13}}} = -3.073$

$\text{Degree of freedom} = 11 + 13 - 2 = 22$

Now,

$t_{critical} \text{ at 0.05 level of significance, 22 degree of freedom } = -1.717$

Since,

$t_{stat} < t_{critical}$

We fail to accept the null hypothesis and reject the null hypothesis. We accept the alternate hypothesis.

We conclude that the mean score for Group 1 is significantly lower than the mean score for Group 2.

There is enough evidence to claim that population mean 1 is less than population 2 at the 0.05 significance level.

6 0

3 years ago

PLEASE HURRY I HAVE 10 MINUTES AND ILL GIVE 50 POINTS!

stealth61 [152]

Answer:

What Table!?!?

Step-by-step explanation:

3 0

3 years ago

Read 2 more answers

According to a Pew Research Center study, in May 2011, 35% of all American adults had a smart phone (one which the user can use

taurus [48]

Answer:

$z=\frac{0.4 -0.35}{\sqrt{\frac{0.35(1-0.35)}{300}}}=1.82$

$p_v =P(z>1.82)=0.034$

So the p value obtained was a very low value and using the significance level given $\alpha=0.1$ we have $p_v$ so we can conclude that we have enough evidence to reject the null hypothesis

And the best conclusion would be:

There is enough evidence to show that more than 35% of community college students own a smart phone (Pvalue = 0.034).

And for the second case the correct system of hypothesis is:

H0: p = 0.078; Ha: p > 0.078

Step-by-step explanation:

Data given and notation

n=300 represent the random sample taken

$\hat p=\frac{120}{300}=0.4$ estimated proportion of college students that have a smart phone

$p_o=0.35$ is the value that we want to test

$\alpha=0.1$ represent the significance level

Confidence=90% or 0.90

z would represent the statistic (variable of interest)

$p_v$ represent the p value (variable of interest)

Concepts and formulas to use

We need to conduct a hypothesis in order to test the claim that the proportion is >0.35.:

Null hypothesis: $p\leq 0.35$

Alternative hypothesis: $p > 0.35$

When we conduct a proportion test we need to use the z statistic, and the is given by:

$z=\frac{\hat p -p_o}{\sqrt{\frac{p_o (1-p_o)}{n}}}$ (1)

The One-Sample Proportion Test is used to assess whether a population proportion $\hat p$ is significantly different from a hypothesized value $p_o$ .

Calculate the statistic

Since we have all the info requires we can replace in formula (1) like this:

$z=\frac{0.4 -0.35}{\sqrt{\frac{0.35(1-0.35)}{300}}}=1.82$

Statistical decision

It's important to refresh the p value method or p value approach . "This method is about determining "likely" or "unlikely" by determining the probability assuming the null hypothesis were true of observing a more extreme test statistic in the direction of the alternative hypothesis than the one observed". Or in other words is just a method to have an statistical decision to fail to reject or reject the null hypothesis.

The significance level provided $\alpha=0.1$ . The next step would be calculate the p value for this test.

Since is a right tailed test the p value would be:

$p_v =P(z>1.82)=0.034$

So the p value obtained was a very low value and using the significance level given $\alpha=0.1$ we have $p_v$ so we can conclude that we have enough evidence to reject the null hypothesis

And the best conclusion would be:

There is enough evidence to show that more than 35% of community college students own a smart phone (Pvalue = 0.034).

And for the second case the correct system of hypothesis is:

H0: p = 0.078; Ha: p > 0.078

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3 years ago

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yan [13]

Answer:

from top to bottom = c, d, a, b

Step-by-step explanation:

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3 years ago

I need to know the square root answer of this problem simplified

Irina18 [472]

12 sqrt3 * 2 sqrt2 * 3

= 24 sqrt6 * 3

= 72 sqrt6 answer

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3 years ago