A family purchased tickets for a concert. They paid a total of $90 for 7 adult tickets and 12 child tickets. The price of a chil
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Answer:
So the p value obtained was a very low value and using the significance level asumed we have
so we can conclude that we have enough evidence to FAIl to reject the null hypothesis, and we can said that at 5% of significance the proportion of women who complain of nausea between the 24th and 28th week of pregnancy is not significantly higher than 0.3 or 30%
Step-by-step explanation:
Data given and notation
n=195 represent the random sample taken
X=65 represent the women who complain of nausea between the 24th and 28th week of pregnancy
estimated proportion of women who complain of nausea between the 24th and 28th week of pregnancy
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)
Concepts and formulas to use
We need to conduct a hypothesis in order to test the claim that true proportion is higher than 0.3.:
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
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Calculate the statistic
Since we have all the info requires we can replace in formula (1) like this:
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 assumed is . The next step would be calculate the p value for this test.
Since is a right tailed test the p value would be:
So the p value obtained was a very low value and using the significance level asumed we have
so we can conclude that we have enough evidence to FAIl to reject the null hypothesis, and we can said that at 5% of significance the proportion of women who complain of nausea between the 24th and 28th week of pregnancy is not significantly higher than 0.3 or 30%
Answer:
gives the solution
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gives the solution
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Step-by-step explanation:
I will solve both interpretations.
If we assume the equation is , then the following is the process:
Add 5 on both sides:
Simplify:
Now write an equivalent logarithm form:
Now using the change of base:
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If we assume the equation is , then we use the following process:
Write an equivalent logarithm form:
Add 5 on both sides:
Use change of base formula: