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Sveta_85 [38]
3 years ago
9

Mrs. McDonald's students have been working very hard during class, completing all class work and homework, and even attending tu

torials after school. The day before the test, a new student was enrolled in the class and Mrs. McDonald had this student take the test to see what they knew. The test scores are listed below. Which of the following would be the best measure of central tendency?

Mathematics
1 answer:
Natali5045456 [20]3 years ago
3 0

Answer:

(13+83+85+87+90+91+93+97+98+99+100+100):12= 85,33

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In a recent study, researchers found that 31 out of 150 boys aged 7-13 were overweight or obese. On the basis of this study can
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Answer:

1) Null hypothesis:p \leq 0.15  

Alternative hypothesis:p > 0.15  

2) The One-Sample Proportion Test is used to assess whether a population proportion \hat p is significantly different from a hypothesized value p_o.

3) z_{crit}= 1.64

And the rejection zone would be z>1.64

4) Calculate the statistic  

z=\frac{0.207 -0.15}{\sqrt{\frac{0.15(1-0.15)}{150}}}=1.955  

5) Statistical decision  

For this case our calculated value is on the rejection zone, so we have enough evidence to reject the null hypothesis at 5% of significance and we can conclude that the true proportion is higher than 0.15

Step-by-step explanation:

Data given and notation

n=150 represent the random sample taken

X=21 represent the boys overweight

\hat p=\frac{31}{150}=0.207 estimated proportion of boy overweigth

p_o=0.15 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)  

1) Concepts and formulas to use  

We need to conduct a hypothesis in order to test the true proportion of boys obese is higher than 0.15.:  

Null hypothesis:p \leq 0.15  

Alternative hypothesis:p > 0.15  

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)  

2) The One-Sample Proportion Test is used to assess whether a population proportion \hat p is significantly different from a hypothesized value p_o.

3) Decision rule

For this case we need a value on the normal standard distribution who accumulates 0.05 of the area on the right tail and on this case this value is:

z_{crit}= 1.64

And the rejection zone would be z>1.64

4) Calculate the statistic  

Since we have all the info requires we can replace in formula (1) like this:  

z=\frac{0.207 -0.15}{\sqrt{\frac{0.15(1-0.15)}{150}}}=1.955  

5) Statistical decision  

For this case our calculated value is on the rejection zone, so we have enough evidence to reject the null hypothesis at 5% of significance and we can conclude that the true proportion is higher than 0.15

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