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A publisher reports that 65% of their readers own a laptop. A marketing executive wants to test the claim that the percentage is

muminat

Answer:

$z=\frac{0.60 -0.65}{\sqrt{\frac{0.65(1-0.65)}{340}}}=-1.933$

The p value for this case can be calculated with this probability:

$p_v =2*P(z$

For this case is we use a significance level of 5% we have enough evidence to FAIL to reject the null hypothesis and we can't conclude that the true proportion is different from 0.65 or 65%. We need to be careful since if we use a value higher than 65 for the significance the result would change

Step-by-step explanation:

Information given

n=340 represent the random sample taken

$\hat p=0.60$ estimated proportion of readers owned a laptop

$p_o=0.65$ is the value that we want to test

z would represent the statistic

$p_v{/tex} represent the p valueHypothesis to testWe want to check if the true proportion of readers owned a laptop if different from 0.65
Null hypothesis:[tex]p=0.65$

Alternative hypothesis: $p \neq 0.65$

The statistic is given by:

$z=\frac{\hat p -p_o}{\sqrt{\frac{p_o (1-p_o)}{n}}}$ (1)

Replacing we got:

$z=\frac{0.60 -0.65}{\sqrt{\frac{0.65(1-0.65)}{340}}}=-1.933$

The p value for this case can be calculated with this probability:

$p_v =2*P(z$

For this case is we use a significance level of 5% we have enough evidence to FAIL to reject the null hypothesis and we can't conclude that the true proportion is different from 0.65 or 65%. We need to be careful since if we use a value higher than 65 for the significance the result would change

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4 years ago

How do I do Q43B someone please help!!

Setler79 [48]

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Step2247 [10]

I would answer the problem but first upload the diagram

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