a. We fail to reject the hypothesis that the average credit score for an adult in Virginia is different from the average credit score for an adult in North Carolina at the significance level of 0.05.

b. The 95% confidence interval for the true difference of means is -2.2468 and 36.2468. There is a probability of 95% that the true difference of means $\mu_{1}-\mu_{2}$ is between -2.2468 and 36.2468. This confidence interval contains the number 0, this is consistent with the results we got in a. Because we fail to reject the hypothesis that the average credit score for an adult in Virginia is different from the average credit score for an adult in North Carolina at the significance level of 0.05.

Step-by-step explanation:

Let $\mu_{1}-\mu_{2}$ be the true difference between the average credit score for an adult in Virginia and the average credit score for an adult in North Carolina. We have the large sample sizes $n_{1} = 40$ and $n_{2} = 35$ , the unbiased point estimate for $\mu_{1}-\mu_{2}$ is $\bar{x}_{1} - \bar{x}_{2}$ , i.e., 699-682 = 17.

The standard error is given by $\sqrt{\frac{\sigma^{2}_{1}}{n_{1}}+\frac{\sigma^{2}_{2}}{n_{2}}}$ , i.e.,

$\sqrt{\frac{(44)^{2}}{40}+\frac{(41)^{2}}{35}}$ = 9.8198.

a. We want to test $H_{0}: \mu_{1}-\mu_{2} = 0$ vs $H_{1}: \mu_{1}-\mu_{2} \neq 0$ (two-tailed alternative). The rejection region is given by RR = {z | z < -1.96 or z > 1.96} where -1.96 and 1.96 are the 2.5th and 97.5th quantiles of the standard normal distribution respectively. The test statistic is $Z = \frac{\bar{x}_{1} - \bar{x}_{2}-0}{\sqrt{\frac{\sigma^{2}_{1}}{n_{1}}+\frac{\sigma^{2}_{2}}{n_{2}}}}$ and the observed value is $z_{0} = \frac{17}{9.8198} = 1.7312$ . Because 1.7312 does not fall inside RR, we fail to reject the null hypothesis.

b. The endpoints for a 95% confidence interval for $\mu_{1}-\mu_{2}$ is given by $17\pm (z_{0.05/2})9.8198$ , i.e., $17\pm (z_{0.025})9.8198$ where $z_{0.025}$ is the 2.5th quantile of the standard normal distribution, i.e., -1.96, so, we have 17-(1.96)(9.8198) and 17+(1.96)(9.8198), i.e., -2.2468 and 36.2468. There is a probability of 95% that the true difference of means $\mu_{1}-\mu_{2}$ is between -2.2468 and 36.2468. This confidence interval contains the number 0, this is consistent with the results we got in a. Because we fail to reject the hypothesis that the average credit score for an adult in Virginia is different from the average credit score for an adult in North Carolina at the significance level of 0.05.

3 0

3 years ago

If the area of a triangular kite is 60 square feet and its base is 10 feet, find the height of the kite.

11Alexandr11 [23.1K]

The area of the triangular kite would be 12 feet

3 0

2 years ago

PLEASE HELP I WILL MARK BRANLIEST

Finger [1]

Answer:

SAS

Step-by-step explanation:

We must prove that triangles ABC and EDC are congruent.

Since BD bisects AE, then AC is congruent with CE.

Since AE bisects BD, then BC is congruent with CD

Angle C is 90° in the triangle EDC and is also 90° in triangle BCA because they are vertical angles.

Being two sides and the included angle congruent, then both triangles are similar by the SAS theorem.

Answer: SAS

4 0

2 years ago

9. The endpoints of a diameter of a circle are A (2,1) and B(5,5). Find the area of the circle in terms of . (1 point)

Georgia [21]

Answer:

Area of the circle is 6.25π sq.unit

Step-by-step explanation:

The centre of the circle is at the mid point of a diameter. Th mid point of the diameter is at (5+22,5+12)or(72,3),which is centre. Radius is the distance from end point (5,5)to centre (72,3)∴r=√(5−72)2+(5−3)2=√1.52+4=√6.25=2.5∴

Area of the circle is Ac=π⋅r2=π⋅2.52=6.25π sq.unit[Ans]

5 0

2 years ago

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