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Tpy6a [65]
3 years ago
8

There are 63 students in a speech contest.Yesterday 1/3 gave a speech and today 2/3 of the remaining students gave their speech.

how many students still haven't given their speeches​
Mathematics
1 answer:
Paraphin [41]3 years ago
7 0

Hey!

-----------------------------------------------

Solution:

Find 1/3 of 63.

63 / 3 = 21

21 students gave a speech today.

63 - 21 = 42

Find 2/3 of 42.

42 / 3 = 14

14 x 2 = 28

28 of the remaining students gave their speeches.

42 - 28 = 14

-----------------------------------------------

Answer:

14 students haven't given their speeches!

-----------------------------------------------

Hope This Helped! Good Luck!

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Answer:

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

z=\frac{0.34848-0.34782}{\sqrt{0.34831(1-0.34831)(\frac{1}{66}+\frac{1}{23})}}=0.0057  

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The p value is a very high value and using the significance level \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 percent of men who enjoy shopping for electronic equipment is NOT significantly higher than the percent of women who enjoy shopping for electronic equipment.  

Step-by-step explanation:

1) Data given and notation  

X_{M}=23 represent the number of men that said they enjoyed the activity of Saturday afternoon shopping

X_{W}=8 represent the number of women that said they enjoyed the activity of Saturday afternoon shopping

n_{M}=66 sample of male selected

n_{W}=23 sample of demale selected

p_{M}=\frac{23}{66}=0.34848 represent the proportion of men that said they enjoyed the activity of Saturday afternoon shopping

p_{W}=\frac{8}{23}=0.34782 represent the proportion of women with red/green color blindness  

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 percent of men who enjoy shopping for electronic equipment is higher than the percent of women who enjoy shopping for electronic equipment , the system of hypothesis would be:  

Null hypothesis:p_{M} \leq 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{23+8}{66+23}=0.34831

3) Calculate the statistic

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

z=\frac{0.34848-0.34782}{\sqrt{0.34831(1-0.34831)(\frac{1}{66}+\frac{1}{23})}}=0.0057  

4) Statistical decision

Using the significance level provided \alpha=0.05, the next step would be calculate the p value for this test.  

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

p_v =P(Z>0.0057)=0.4977  

So the p value is a very high value and using the significance level \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 percent of men who enjoy shopping for electronic equipment is NOT significantly higher than the percent of women who enjoy shopping for electronic equipment.  

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