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KATRIN_1 [288]
4 years ago
10

Beth spent 3 hours 15 mins writing a paper and 30 mins taking a break. She finished at 7:20 pm. What time did she start?

Mathematics
1 answer:
svp [43]4 years ago
3 0

Answer:

a

Step-by-step explanation:

hope this helps

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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

5 0
4 years ago
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