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Andreyy89
3 years ago
8

Unit of Measurement, a punk/disco band, played a benefit concert for New Year's Eve. They started at 7:00 p.m. on New Year's Eve

and managed to play until 8:30 a.m. the following morning. How many minutes did they play in all?
Mathematics
1 answer:
Maksim231197 [3]3 years ago
6 0
810
count the number of hours they played which should be 13 hrs and 30 mins,
then since every hour is 60 mins multiply 60 by 13 and then add 30
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In a random sample of males, it was found that 28 write with their left hands and 210 do not. In a random sample of females, it
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Answer:

a) Null hypothesis:p_{M} \geq p_{W}  

Alternative hypothesis:p_{M} < p_{W}  

b) z=\frac{0.118-0.125}{\sqrt{0.122(1-0.122)(\frac{1}{238}+\frac{1}{497})}}=-0.271    

c)p_v =P(Z   

d) If we compare the p value and the significance level given \alpha=0.05 we see that p_v>\alpha so we can conclude that we have enough evidence to FAIL to reject the null hypothesis, and we can say the the proportion of men that write with their left hand is NOT significant lower than the proportion of female that write with their left hand

Step-by-step explanation:

1) Data given and notation  

X_{M}=28 represent the number of men that write with their left hand

X_{W}=62 represent the number of women  that write with their left hand

n_{M}=210+28=238 sample of male selected

n_{W}=62+435=497 sample of female selected

p_{M}=\frac{28}{238}=0.118 represent the proportion of men  that write with their left hand

p_{W}=\frac{62}{497}=0.125 represent the proportion of women that write with their left hand

\alpha=0.05 represent the significance level

z would represent the statistic (variable of interest)  

p_v represent the value for the test (variable of interest)

2) Concepts and formulas to use  

We need to conduct a hypothesis in order to check if the rate of left-handedness among males is less than that among females , the system of hypothesis would be:  

Null hypothesis:p_{M} \geq p_{W}  

Alternative hypothesis:p_{M} < p_{W}  

We need to apply a z test to compare proportions, and the statistic is given by:  

z=\frac{p_{M}-p_{W}}{\sqrt{\hat p (1-\hat p)(\frac{1}{n_{M}}+\frac{1}{n_{W}})}}   (1)

Where \hat p=\frac{X_{M}+X_{W}}{n_{M}+n_{W}}=\frac{28+62}{238+497}=0.122

3) Calculate the statistic

Replacing in formula (1) the values obtained we got this:  

z=\frac{0.118-0.125}{\sqrt{0.122(1-0.122)(\frac{1}{238}+\frac{1}{497})}}=-0.271  

4) Statistical decision

We have a significance level provided \alpha=0.05, and now we can calculate the p value for this test.  

Since is a one left side test the p value would be:  

p_v =P(Z  

If we compare the p value and the significance level given \alpha=0.05, 0,1,0.15 we see that p_v>\alpha so we can conclude that we have enough evidence to FAIL to reject the null hypothesis, and we can say the the proportion of men that write with their left hand is NOT significant loer than the proportion of female that write with their left hand .  

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3 years ago
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