Relative Frequencies and Association
2 answers:
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<h3>The absolute error is 6</h3><h3>The percent error is 16.22 %</h3>
<em><u>Solution:</u></em>
Given that,
Actual volume of his bathtub is 37 gal
Estimate is 31 gal
<em><u>Find the absolute error</u></em>
Absolute Error = | Measured Value - Actual Value |
Absolute Error = | 31 - 37 |
Absolute Error = 6
<em><u>Find the percent error</u></em>
Substituting we get,
Negative sign means percent decrease
Thus percent error is 16.22 %
Answer:
With the most common significance levels used we see that the p value is lower than the significance level so then we have enough evidence to reject the null hypothesis and we can say that the true proportion is significantly higher than 0.5
Step-by-step explanation:
Information given
n=297 represent the random sample of male taken
X=183 represent the men who said yes, they had driven a car when they probably had too much alcohol
estimated proportion of men who said yes, they had driven a car when they probably had too much alcohol
is the value that we want to test
z would represent the statistic (variable of interest)
represent the p value (variable of interest)
Hypothesis to test
We need to conduct a hypothesis in order to test the claim that the majority of men in the population (that is, more than half) would say that they had driven a car when they probably had too much alcohol, and the system of hypothesis are:
Null hypothesis:
Alternative hypothesis:
The statistic is given by:
(1)
After replace we got:
Since we have all the info requires we can replace in formula (1) like this:
Decision
We have a right tailed test so then the p value would be:
With the most common significance levels used we see that the p value is lower than the significance level so then we have enough evidence to reject the null hypothesis and we can say that the true proportion is significantly higher than 0.5
The answer is D
To solve the answer you would have needed to multiply the reciprocal of but in this case, Unvi multiplied with the reciprocal of 14.
Answer:
D)Yes, because the difference in the means in the actual experiment was more than two standard deviations from 0.
Step-by-step explanation:
We will test the hypothesis on the difference between means.
We have a sample 1 with mean M1=18.2 (drug group) and a sample 2 with mean M2=15.9 (no-drug group).
Then, the difference between means is:
If the standard deviation of the differences of the sample means of the two groups was 1.1 days, the t-statistic can be calculated as:
The critical value for a two tailed test with confidence of 95% (level of significance of 0.05) is t=z=1.96, assuming a large sample.
This is approximately 2 standards deviation (z=2).
The test statistict=2.09 is bigger than the critical value and lies in the rejection region, so the effect is significant. The null hypothesis would be rejected: the difference between means is significant.