(23/6)*(6/4)*(7/5)=?
2 answers:
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Answer:
15, 8, 66%, 40
Step-by-step explanation:
Hi! ok so in ogive charts its really about finding the corresponding data points and sometimes it's a lot of estimation so these answers may not be exact but they'll be close.
In the image, you'll see a bunch of lines, basically what youre doing is finding the corresponding frequency of the dollars or vice versa.
<em>question a)</em>
50 dollars or less, so the area of 50 dollars and below that.
from the chart we can see that this lines up to around 15 people.
<em>question b)</em>
now this one is a little bit more challenging, but you can use the same process as you did for the first question. so we can see who spent 100 dollars or less. this value is around 42 (just from looking at the graph)
<em>we know if you didn't spend 100 dollars or less, you must have spent more than that. </em>
so take 50-42=8
around 8<u> people</u> spent 100 dollars or more.
<em>question c)</em>
basically the same question as the first but more calculations are involved. from the graph we can see around 33 people spent 80 dollars or less. but the question is asking the percentage. so what percent of 50 (people) is 33.
33/50 = 0.66
around 66%
<em>question d)</em>
80th percentile means less than or equal to 80% of the frequencies/the data. so we first need to find what value that is in terms of frequency.
50 times 0.8 = 40
40 people
now just plot that on the graph (sorry i did it wrong in the image)
Answer:
If we compare the p value obtained and the significance level given we have
so we can conclude that we have enough evidence to reject the null hypothesis, and we can said that at 5% of significance the proportion of students reported experiencing excessive daytime sleepiness (EDS) is significantly higher than 0.5 or the half.
Step-by-step explanation:
1) Data given and notation
n=377 represent the random sample taken
X=209 represent the students reported experiencing excessive daytime sleepiness (EDS)
estimated proportion of students reported experiencing excessive daytime sleepiness (EDS)
is the value that we want to test
represent the significance level
Confidence=95% or 0.95
z would represent the statistic (variable of interest)
represent the p value (variable of interest)
2) Concepts and formulas to use
We need to conduct a hypothesis in order to test the claim that the true proportion is higher than 0.5:
Null hypothesis:
Alternative hypothesis:
When we conduct a proportion test we need to use the z statistic, and the is given by:
(1)
The One-Sample Proportion Test is used to assess whether a population proportion is significantly different from a hypothesized value
.
3) Calculate the statistic
Since we have all the info requires we can replace in formula (1) like this:
4) Statistical decision
It's important to refresh the p value method or p value approach . "This method is about determining "likely" or "unlikely" by determining the probability assuming the null hypothesis were true of observing a more extreme test statistic in the direction of the alternative hypothesis than the one observed". Or in other words is just a method to have an statistical decision to fail to reject or reject the null hypothesis.
The significance level provided . The next step would be calculate the p value for this test.
Since is a right tailed test the p value would be:
If we compare the p value obtained and the significance level given we have
so we can conclude that we have enough evidence to reject the null hypothesis, and we can said that at 5% of significance the proportion of students reported experiencing excessive daytime sleepiness (EDS) is significantly higher than 0.5 or the half.