Mr.sanches drives 110 miles in 3 hours. At the same rate, how far will he drive in 5hours

2 answers:

8 0

$\bf \begin{array}{ccll} miles&hours\\ \cline{1-2} 110&3\\ x&5 \end{array}\implies \cfrac{110}{x}=\cfrac{3}{5}\implies 550=3x\implies \cfrac{550}{3}=x\implies 183\frac{1}{3}=x$

kozerog [31]3 years ago

3 0

Answer:

550

Step-by-step explanation:

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A television network is deciding whether or not to give its newest television show a spot during prime viewing time at night. If

REY [17]

Answer:

b. 1.71

$z=\frac{0.5296 -0.5}{\sqrt{\frac{0.5(1-0.5)}{827}}}=1.71$

Step-by-step explanation:

1) Data given and notation

n=827 represent the random sample taken

X=438 represent the people that indicate that they would like to see the new show in the lineup

$\hat p=\frac{438}{827}=0.5296$ estimated proportion of people that indicate that they would like to see the new show in the lineup

$p_o=0.5$ is the value that we want to test

$\alpha$ represent the significance level

z would represent the statistic (variable of interest)

$p_v$ represent the p value

2) Concepts and formulas to use

We need to conduct a hypothesis in order to test the claim that the true proportion is higher than 0.5:

Null hypothesis: $p \leq 0.5$

Alternative hypothesis: $p > 0.5$

When we conduct a proportion test we need to use the z statisitc, 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:

$z=\frac{0.5296 -0.5}{\sqrt{\frac{0.5(1-0.5)}{827}}}=1.71$

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

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

$p_v =P(Z>1.71)=0.044$

If we compare the p value obtained and the significance level assumed $\alpha=0.05$ 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 true proportion is higher than 0.5.