1 1/2 as mixed number

In a survey conducted by a website, employers were asked if they had ever sent an employee home because they were dressed inappr

castortr0y [4]

Answer:

$z=\frac{0.353 -0.333}{\sqrt{\frac{0.333(1-0.333)}{2759}}}=2.23$

$p_v =P(z>2.23)=0.0129$

So the p value obtained was a very low value and using the significance level given $\alpha=0.05$ we have $p_v$ so we can conclude that we have enough evidence to reject the null hypothesis, and we can said that at 5% of significance the proportion of more than one-third of employers have sent an employee home to change clothes is higher than 1/3 or 0.333.

Step-by-step explanation:

Data given and notation

n=2759 represent the random sample taken

X=974 represent the people saying that they had sent an employee home for inappropriate attire

$\hat p=\frac{974}{2759}=0.353$ estimated proportion of people saying that they had sent an employee home for inappropriate attire

$p_o=0.333$ is the value that we want to test

$\alpha=0.05$ represent the significance level

Confidence=95% or 0.95

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 more than one-third of employers have sent an employee home to change clothes.:

Null hypothesis: $p \leq 0.33$

Alternative hypothesis: $p > 0.33$

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:

$z=\frac{0.353 -0.333}{\sqrt{\frac{0.333(1-0.333)}{2759}}}=2.23$

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 $\alpha=0.05$ . The next step would be calculate the p value for this test.

Since is a right tailed the p value would be:

$p_v =P(z>2.23)=0.0129$

So the p value obtained was a very low value and using the significance level given $\alpha=0.05$ we have $p_v$ so we can conclude that we have enough evidence to reject the null hypothesis, and we can said that at 5% of significance the proportion of more than one-third of employers have sent an employee home to change clothes is higher than 1/3 or 0.333.

4 0

3 years ago

I have been stuck on this all day, please explain how to do this

german

Answer:$3,700

Step-by-step explanation:

Gross sales are the total amount of sales a company earned throughout a specific period of time, without taking into consideration any costs involved with running a business. Gross sales do not factor in expenses related to running a business, also known as cost of goods sold (COGS), which get deducted when calculating net sales. For example, they do not account for costs associated with item production, employee wages, building rent, returns, theft or sales tax.

Gross Sales = $3,700

8 0

3 years ago

The radius of a circle is measured at 15.6cm. The actual radius is 15.3cm. Find, to the nearest percent, the percent error in th

Anon25 [30]

Answer:

$Percentage\hspace{3}error\approx 1.96\%$

Step-by-step explanation:

The error percentage is a measure of how inaccurate a measurement is, standardized based on the size of the measurement. It can be easily calculated using the following formula:

$Percentage\hspace{3}error=|\frac{v_A-v_E}{v_E} | \times 100$

Where:

$v_A=Approximate\hspace{3}value\\v_E=Exact\hspace{3}value$

Therefore, according to the data provided by the problem:

$v_A=15.6\\v_E=15.3$

The percentage error is:

$Percentage\hspace{3}error=|\frac{15.6-15.3}{15.3}| \times 100 = 1.960784314\%\approx 1.96\%$

8 0

4 years ago

For the given hypothesis test, determine the probability of a Type II error or the power, as specified. A hypothesis test is to

erica [24]

Answer:

the probability of a Type II error if in fact the mean waiting time u, is 9.8 minutes is 0.1251

Option A) is the correct answer.

Step-by-step explanation:

Given the data in the question;

we know that a type 11 error occur when a null hypothesis is false and we fail to reject it.

as in it in the question;

obtained mean is 9.8 which is obviously not equal to 8.3

But still we fail to reject the null hypothesis says mean is 8.3

Hence we have to find the probability of type 11 error

given that; it is right tailed and o.5, it corresponds to 1.645

so

z is equal to 1.645

z = (x-μ)/ $\frac{S}{\sqrt{n} }$

where our standard deviation s = 3.8

sample size n = 50

mean μ = 8.3

we substitute

1.645 = (x - 8.3)/ $\frac{3.8}{\sqrt{50} }$

1.645 = (x - 8.3) / 0.5374

0.884023 = x - 8.3

x = 0.884023 + 8.3

x = 9.18402

so, by general rule we will fail to reject the null hypothesis when we will get the z value less than 1.645

As we reject the null hypothesis for right tailed test when the obtained test statistics is greater than the critical value

so, we will fail to reject the null hypothesis as long as we get the sample mean less than 9.18402

Now, for mean 9.8 and standard deviation 3.8 and sample size 50

Z = (9.18402 - 9.8)/ $\frac{3.8}{\sqrt{50} }$

Z = -0.61598 / 0.5374

Z = - 1.1462 ≈ - 1.15

from the z-score table;

P(z<-1.15) = 0.1251

Therefore, the probability of a Type II error if in fact the mean waiting time u, is 9.8 minutes is 0.1251

Option A) is the correct answer.

8 0

3 years ago

Pls help me I am freaking out. :(

kkurt [141]

Andre : Anything less than -100, for example -150

-150<-100

|-150|=150

Han: Anything between -100 & 100, for example 50

50> 100

|50|= 50

Lin: 100 or -100,

100> -100

|100|=100

or

-100=-100

|-100|=100

8 0

3 years ago