According to projections through the year 2030, the population y of the given state in year x is approximated by State A: minu
1 answer:
Answer:
In year 2026 population of both the states will be same.
Step-by-step explanation:
It has been given in the question that the population y of State A in year x will be represented by the equation
-3x + y = 11,400
⇒ y = 11,400 + 3x --------(1)
Similarly, equation that represents the population for State B is
-138x + y = 8,000
⇒ y = 138x + 8,000 -------(2)
If population y for both the states are equal then
3x + 11,400 = 138x + 8,000
11400 - 8000 = 138x - 3x
135x = 3400
x = 25.19 [After 25 years ]
Since x = 0 represents the population of the states in year 2000
therefore, population y will be same in both the states in year (2000 + 26)
= 2026
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Answer:
Critical value is t = 1.9901
Step-by-step explanation:
We are given the results of the sampling :
In-House Credit Card National Credit Card <u>Sample Size</u> : 32 50
<u>Mean Monthly Purchases </u>: $45.67 $39.87
<u>Standard Deviation </u>: $10.90 $12.47
Also, the managers wished to test whether there is a statistical difference in the mean monthly purchases by customers using the two types of credit cards, using a significance level α of 0.05.
<em>Firstly, we will specify our null and alternate hypothesis;</em>
Let = Mean Monthly Purchases of In-House Credit Card
= Mean Monthly purchases of National Credit Card
So, Null Hypothesis, :
= 0 {means that there is no difference in the mean monthly purchases by customers using the two types of credit cards}
Alternate Hypothesis, :
0 {means that there is statistical difference in the mean monthly purchases by customers using the two types of credit cards}
The test statistics that will be used here is <u>Two-sample t-test statistics</u>;
T.S. = ~
where, = Sample mean Purchases of In-House Credit Card = $45.67
= Sample mean Purchases of National Credit Card = $39.87
= pooled variance
= In-house credit card sample = 32
= National credit card sample = 50
So, degree of freedom of t-value here is (32 + 50 - 2) = 80
Now, at 0.05 significance level, t table gives critical value of t = 1.9901 at 80 degree of freedom.
<em>Therefore, the critical value assuming the population standard deviations are not known but that the populations are normally distributed with equal variances is t = 1.9901.</em>