What is the rate of change and initial value for the linear relation that includes the points shown in the table?
1 answer:
The initial value: 25, rate of change: -5 ⇒ answer C
Step-by-step explanation:
The table:
→ x : 1 : 3 : 5 : 7
→ y : 20 : 10 : 0 : -10
The linear relation between x and y is y = m x + b, where
- m is the rate of change
- b is the initial value
Let us use the table to find m and b
∵ , where
and are any two ordered pairs in the table
∵ = (1 , 20)
∵ = (3 , 10)
∴
∵ m is the rate of change
∴ The rate of change is -5
Substitute the value of m in the form of the linear relation
∴ y = -5 x + b
To find b substitute x and y in the equation by the coordinates of any points in the table, let us chose point (5 , 0)
∵ x = 5 and y = 0
∴ 0 = -5(5) + b
∴ 0 = -25 + b
- Add 25 to both sides
∴ 25 = b
∵ b represents the initial value
∴ The initial value is 25
The initial value: 25, rate of change: -5
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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).