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zubka84 [21]
3 years ago
13

Brody and Amanda can you for one hour and 20 minutes before serving to fish at 1:50 PM at what time did they start canoeing

Mathematics
2 answers:
aev [14]3 years ago
6 0
11:55A.M witch means the awnser is A
frozen [14]3 years ago
5 0
11:55, which is the answer A.
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a restaurant bill was $24 after tax. if you wanted to leave a 10% tip how much should you leave total?
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Help plz <br> If SU = 110, find the value of x. Then find ST and TU.<br> 2x - 10<br> 5x + 50
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Answer:

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Step-by-step explanation:

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In a​ survey, 32​% of the respondents stated that they talk to their pets on the telephone. A veterinarian believed this result
adoni [48]

Answer:

z=\frac{0.29 -0.32}{\sqrt{\frac{0.32(1-0.32)}{210}}}=-0.932  

p_v =P(z  

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 proportion of  pet owners contacted by telephone is NOT significantly lower than 0.32

Step-by-step explanation:

Data given and notation

n=210 represent the random sample taken

X=61 represent the number of pet owners contacted by telephone

\hat p=\frac{61}{210}=0.29 estimated proportion of  pet owners contacted by telephone

p_o=0.32 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 the true proportions is lower than 0.32.:  

Null hypothesis:p \geq 0.32  

Alternative hypothesis:p < 0.32  

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{0.29 -0.32}{\sqrt{\frac{0.32(1-0.32)}{210}}}=-0.932  

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 left tailed test the p value would be:  

p_v =P(z  

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 proportion of  pet owners contacted by telephone is NOT significantly lower than 0.32

3 0
3 years ago
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