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1 answer:
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Answer:
The p-value of the test is 0.0041 < 0.05, which means that there is sufficient evidence at the 0.05 significance level to conclude that the mean cost has increased.
Step-by-step explanation:
The average undergraduate cost for tuition, fees, and room and board for two-year institutions last year was $13,252. Test if the mean cost has increased.
At the null hypothesis, we test if the mean cost is still the same, that is:
At the alternative hypothesis, we test if the mean cost has increased, that is:
The test statistic is:
We have the standard deviation for the sample, which means that the t-distribution is used to solve this question
In which X is the sample mean, is the value tested at the null hypothesis, s is the standard deviation and n is the size of the sample.
13252 is tested at the null hypothesis:
This means that
The following year, a random sample of 20 two-year institutions had a mean of $15,560 and a standard deviation of $3500.
This means that
Value of the test statistic:
P-value of the test and decision:
The p-value of the test is found using a t-score calculator, with a right-tailed test, with 20-1 = 19 degrees of freedom and t = 2.95. Thus, the p-value of the test is 0.0041.
The p-value of the test is 0.0041 < 0.05, which means that there is sufficient evidence at the 0.05 significance level to conclude that the mean cost has increased.
Answer:
Th second condition is satisfied but the first one on this case it's not satisfied. So for this case it's not good apply the normal approximation to the distribution of p.
Step-by-step explanation:
We need to check the conditions in order to use the normal approximation.
Th second condition is satisfied but the first one on this case it's not satisfied. So for this case it's not good apply the normal approximation to the distribution of p.
If we have both conditions satisfied the general procedure is the following:
Data given and notation
n=23 represent the random sample taken
estimated proportion
is the value that we want to test
represent the significance level
Confidence=90% or 0.90
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 the true proportion is less than 0.11.:
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
.
Calculate the statistic
On this case since the normal approximation it's not satisfid it's not correct calculate the statistic.
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 left tailed test the p value would be:
And if the we reject the null hypothesis. Otherwise w fail to the reject the null hypothesis.