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Wittaler [7]
3 years ago
15

A total of 715 tickets were sold for the school play. They were either adult tickets or student tickets. There were 65 more stud

ent tickets sold than adult tickets.
Mathematics
2 answers:
Svetlanka [38]3 years ago
7 0

Answer:

adults=325& students=390

Step-by-step explanation:

let me the tickets for adults be x

for students will be x+65

x+x+65=715

2x+65=715

2x=715-65

2x=650

x=650÷2

x=325

Sergeeva-Olga [200]3 years ago
6 0

Answer:

There were 423 adult tickets sold.

Step-by-step explanation:

Let x = the number of adult tickets sold

Let x + 65 = the number of student tickets sold

x + x + 65 = 715

2x = 715

x = 357.5

x + 65 = 422.5 (round up)

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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
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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.  

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