When comparing prices for activities at a local recreation center, you notice that the center charges different prices on differ
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Answer:
The p value obtained and using the significance level assumed we have
so we can conclude that we have enough evidenc to reject the null hypothesis, and we can said that at 5% of significance the proportion of new laptop's with fully charge for the new model is significantly higher compared to the old model.
Step-by-step explanation:
1) Data given and notation n
n=100 represent the random sample taken
X=96 represent the laptop's arrived with fully charged batteries.
estimated proportion of laptop's arrived with fully charged batteries.
is the value that we want to test
represent the significance level
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 new model’s rate is at least as high as the previous model.:
Null hypothesis:
Alternative hypothesis:
When we conduct a proportion test we need to use the z statisitc, 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
P value method or p value approach . "This method consists on 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 is not provided, but we can assume . The next step would be calculate the p value for this test.
Since is a one side test the p value would be:
So based on the p value obtained and using the significance level assumed we have
so we can conclude that we have enough evidenc to reject the null hypothesis, and we can said that at 5% of significance the proportion of new laptop's with fully charge for the new model is significantly higher compared to the old model.
From the dot plot, we have that 90.91% of the students have 5 or more vowels in their first and last names.
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The dot plot states that:
- 2 students have 4 vowels in their first and last name.
- 3 students have 5 vowels.
- 8 students have 6 vowels.
- 9 students have 7 vowels.
Then:
- Total of 2 + 3 + 8 + 9 = 22 students.
- 3 + 8 + 9 = 20 have or more vowels.
90.91% of the students have 5 or more vowels in their first and last names.
A similar problem is given at brainly.com/question/14354536