Wayne bought blueberries. He uses 3/8 of the blueberries to make blueberry bread, 1/6 of the blueberries to make pancakes, and
2 answers:
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Answer:
Step-by-step explanation:
1) Data given and notation
n=100 represent the random sample taken
X=18 represent the ounce cups of coffee that were underfilled
estimated proportion of ounce cups of coffee that were underfilled
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 true proportion it's higher than 0.1 or 10%:
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 assumed for this case is . The next step would be calculate the p value for this test.
Since is a one right tailed test the p value would be:
So the p value obtained was a very low value and using 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 thetrue proportion is not significanlty higher than 0.1 or 10% .
Answer:
No, there is not enough evidence at the 0.02 level to support the executive's claim.
Step-by-step explanation:
We are given that a publisher reports that 34% of their readers own a particular make of car. A random sample of 220 found that 30% of the readers owned a particular make of car.
And, a marketing executive wants to test the claim that the percentage is actually different from the reported percentage, i.e;
Null Hypothesis, : p = 0.34 {means that the percentage of readers who own a particular make of car is same as reported 34%}
Alternate Hypothesis, : p
0.34 {means that the percentage of readers who own a particular make of car is different from the reported 34%}
The test statistics we will use here is;
T.S. = ~ N(0,1)
where, p = actual % of readers who own a particular make of car = 0.34
= percentage of readers who own a particular make of car in a
sample of 220 = 0.30
n = sample size = 220
So, Test statistics =
= -1.30
Now, at 0.02 significance level, the z table gives critical value of -2.3263 to 2.3263. Since our test statistics lie in the range of critical values which means it doesn't lie in the rejection region, so we have insufficient evidence to reject null hypothesis.
Therefore, we conclude that the actual percentage of readers who own a particular make of car is same as reported percentage and the executive's claim that it is different is not supported.
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