Expand the expression.<br> 8(x-5)<br><br> 8(x-5)=_x

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

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Step-by-step explanation:

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