Which of the following is the equation of a line that is perpendicular to the graph of y=2/5x-1

In a study of the accuracy of fast food drive-through orders, McDonald’s had 33 orders that were not accurate among 362 orders o

melomori [17]

Answer:

A. We need to conduct a hypothesis in order to test the claim that the true proportion of inaccurate orders p is 0.1.

B. Null hypothesis: $p=0.1$

Alternative hypothesis: $p \neq 0.1$

C. $z=\frac{0.0912 -0.1}{\sqrt{\frac{0.1(1-0.1)}{362}}}=-0.558$

D. $z_{\alpha/2}=-1.96$ $z_{1-\alpha/2}=1.96$

E. Fail to the reject the null hypothesis

F. So the p value obtained was a very high value and using the significance level given $\alpha=0.05$ we have $p_v>\alpha$ so we can conclude that we have enough evidence to FAIL to reject the null hypothesis, and we can said that at 5% of significance the true proportion of inaccurate orders is not significantly different from 0.1.

Step-by-step explanation:

Data given and notation

n=362 represent the random sample taken

X=33 represent the number of orders not accurate

$\hat p=\frac{33}{363}=0.0912$ estimated proportion of orders not accurate

$p_o=0.10$ 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)

A: Write the claim as a mathematical statement involving the population proportion p

We need to conduct a hypothesis in order to test the claim that the true proportion of inaccurate orders p is 0.1.

B: State the null (H0) and alternative (H1) hypotheses

Null hypothesis: $p=0.1$

Alternative hypothesis: $p \neq 0.1$

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 $p_o$ .

C: Find the test statistic

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

$z=\frac{0.0912 -0.1}{\sqrt{\frac{0.1(1-0.1)}{362}}}=-0.558$

D: Find the critical value(s)

Since is a bilateral test we have two critical values. We need to look on the normal standard distribution a quantile that accumulates 0.025 of the area on each tail. And for this case we have:

$z_{\alpha/2}=-1.96$ $z_{1-\alpha/2}=1.96$

P value

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 bilateral test the p value would be:

$p_v =2*P(z$

E: Would you Reject or Fail to Reject the null (H0) hypothesis.

Fail to the reject the null hypothesis

F: Write the conclusion of the test.

So the p value obtained was a very high value and using the significance level given $\alpha=0.05$ we have $p_v>\alpha$ so we can conclude that we have enough evidence to FAIL to reject the null hypothesis, and we can said that at 5% of significance the true proportion of inaccurate orders is not significantly different from 0.1.

6 0

3 years ago

A random sample of 36 students at a community college showed an average age of 25 years. Assume the ages of all students at the

Pavel [41]

Answer:

98% confidence interval for the average age of all students is [24.302 , 25.698]

Step-by-step explanation:

We are given that a random sample of 36 students at a community college showed an average age of 25 years.

Also, assuming that the ages of all students at the college are normally distributed with a standard deviation of 1.8 years.

So, the pivotal quantity for 98% confidence interval for the average age is given by;

P.Q. = $\frac{\bar X - \mu}{\frac{\sigma}{\sqrt{n} } }$ ~ N(0,1)

where, $\bar X$ = sample average age = 25 years

$\sigma$ = population standard deviation = 1.8 years

n = sample of students = 36

$\mu$ = population average age

So, 98% confidence interval for the average age, $\mu$ is ;

P(-2.3263 < N(0,1) < 2.3263) = 0.98

P(-2.3263 < $\frac{\bar X - \mu}{\frac{\sigma}{\sqrt{n} } }$ < 2.3263) = 0.98

P( $-2.3263 \times {\frac{\sigma}{\sqrt{n} }$ < ${\bar X - \mu}$ < $2.3263 \times {\frac{\sigma}{\sqrt{n} }$ ) = 0.98

P( $\bar X - 2.3263 \times {\frac{\sigma}{\sqrt{n} }$ < $\mu$ < $\bar X +2.3263 \times {\frac{\sigma}{\sqrt{n} }$ ) = 0.98

98% confidence interval for $\mu$ = [ $\bar X - 2.3263 \times {\frac{\sigma}{\sqrt{n} }$ , $\bar X +2.3263 \times {\frac{\sigma}{\sqrt{n} }$ ]

= [ $25 - 2.3263 \times {\frac{1.8}{\sqrt{36} }$ , $25 + 2.3263 \times {\frac{1.8}{\sqrt{36} }$ ]

= [24.302 , 25.698]

Therefore, 98% confidence interval for the average age of all students at this college is [24.302 , 25.698].

8 0

3 years ago

Emilio’s paycheck was $305.25, and he makes $9.25 per hour. How many hours did Emilio work?

Anastaziya [24]

$305.25 divided by $9.25 equals 33, so he worked 33 hours... hope this helped.

5 0

4 years ago

Read 2 more answers

——————————————————————————————

DochEvi [55]

4. 3(3) + 2
(3 x 3) + 2 = 11
5. -(-6) -7
(+)6 -7 = -1

7 0

3 years ago

What is the image point of (-6, -6) after A translation right 2 units and down 3 units

Anton [14]

The point would be (-4,-9) im pretty sure

6 0

3 years ago