Please help will give you brainliest
2 answers:
Answer: Option A
Step-by-step explanation:
Given two variables X and Z, it is said that there is a correlation between X and Z if both variables change together.
That is, if when the variable X increases the variabl Z increases also then there is a positive correlation
if when the variable X increases the variable Z decreases then there is a negative correlation.
Now observe the graphs that are shown in the image. Note that in both cases the points are scattered and there is no clear relationship between the variables. Therefore the correlation is zero in both cases
Examples of strong positive and negative correlations are shown in the attached image
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Answer:
Population parameter(s): mean μ=20
Sample statistics: M=20.7, s=2.1
Hypotheses:
Test Statistic: t=2.357
p-value: 0.011
Reject H? YES
Conclusion (in context of the problem): There is enough evidence to support the claim that the advertised delivery time is overly optimistic and it is larger than 20 minutes.
<em>Is this difference practically significant? </em>No, the difference as the sample mean delivery time is under one minute of difference from the advertised time. Although it is significantly from the statistical point of view, the difference is under 5% of the advertised delivery time.
The 99% confidence interval for the mean is (19.904, 21.496).
With this level of confidence, the null hypothesis failed to be rejected, as t=20 is a possible value for the true mean (is included in the confidence interval).
Step-by-step explanation:
This is a hypothesis test for the population mean.
The claim is that the advertised delivery time is overly optimistic and it is larger than 20 minutes.
Then, the null and alternative hypothesis are:
The significance level is 0.05.
The sample has a size n=50.
The sample mean is M=20.7.
As the standard deviation of the population is not known, we estimate it with the sample standard deviation, that has a value of s=2.1.
The estimated standard error of the mean is computed using the formula:
Then, we can calculate the t-statistic as:
The degrees of freedom for this sample size are:
This test is a right-tailed test, with 49 degrees of freedom and t=2.357, so the P-value for this test is calculated as (using a t-table):
As the P-value (0.011) is smaller than the significance level (0.05), the effect is significant.
The null hypothesis is rejected.
There is enough evidence to support the claim that the advertised delivery time is overly optimistic and it is larger than 20 minutes.
2. We have to calculate a 99% confidence interval for the mean.
The sample mean is M=20.7.
The sample size is N=50.
The t-value for a 99% confidence interval is t=2.68.
The margin of error (MOE) can be calculated as:
Then, the lower and upper bounds of the confidence interval are:
The 99% confidence interval for the mean is (19.904, 21.496).
With this level of confidence, the null hypothesis failed to be rejected, as t=20 is a possible value for the true mean (is included in the confidence interval).
