What is 22% of 60? <br> 13.2<br> 27.3<br> 1,320<br> 38

In a poll of 1000 adults in July 2010, 540 of those polled said that schools should ban sugary snacks and soft drinks. Complete

erastova [34]

Answer:

$z=2.53$

$p_v =P(z>2.53)=0.0057$

The p value obtained was a very low value and using the significance level given $\alpha=0.05$ we have $p_v$ so then we have enough evidence to reject the null hypothesis, and we can say that at 5% of significance, the proportion of adults who support a ban on sugary snacks and soft drinks is more than 0.5 or 50%.

Step-by-step explanation:

1) Data given and notation

n=1000 represent the random sample taken

X=540 represent the adults that said that schools should ban sugary snacks and soft drinks

$\hat p=\frac{540}{1000}=0.54$ estimated proportion of adults that said that schools should ban sugary snacks and soft drinks

$p_o=0.5$ 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)

2) Concepts and formulas to use

We need to conduct a hypothesis in order to test the claim that majority of adults (more than 50%) support a ban on sugary snacks and soft drinks, the system of hypothesis are:

Null hypothesis: $p\leq 0.5$

Alternative hypothesis: $p > 0.5$

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 $p_o$ .

3) Calculate the statistic

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

$z=\frac{0.54 -0.5}{\sqrt{\frac{0.5(1-0.5)}{1000}}}=2.53$

4) 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 one side right tailed test the p value would be:

$p_v =P(z>2.53)=0.0057$

So the p value obtained was a very low value and using the significance level given $\alpha=0.05$ we have $p_v$ so then we have enough evidence to reject the null hypothesis, and we can say that at 5% of significance, the proportion of adults who support a ban on sugary snacks and soft drinks is more than 0.5 or 50%.

7 0

2 years ago

What is the median of 26,38,11,17,102,88,33

Ierofanga [76]

45 is the median of your question

8 0

2 years ago

Read 2 more answers

Write three numbers that when divided by 8 give a negative whole number answer.

abruzzese [7]

-8n where n > 0 and whole number (-8n/8=-n)

-8*1=-8