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kirza4 [7]
1 year ago
5

We suspect that automobile insurance premiums (in dollars) may be steadily decreasing with the driver's driving experience (in y

ears), so we choose a random sample of drivers who have similar automobile insurance coverage and collect data about their ages and insurance premiums. Which of the following is the most appropriate statistical test to use to determine if insurance premiums are decreasing with the driver's driving experience? A. matched pairs t-test
B. two-sample t-test

C. ANOVA

D. chi-squared test for independence

E. inference for regression
Mathematics
1 answer:
ANTONII [103]1 year ago
8 0

Answer:

D) a chi square test for independence.

Step-by-step explanation:

Given that we  suspect that automobile insurance premiums (in dollars) may be steadily decreasing with the driver's driving experience (in years), so we choose a random sample of drivers who have similar automobile insurance coverage and collect data about their ages and insurance premiums.

We are to check whether two variables insurance premiums and driving experience are associated.

Two categorical variables are compared for different ages and insurance premiums.

Hence a proper test would be

D) a chi square test for independence.

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A store ships cans by weight. A small box can hold 3 to 5 pounds. A medium box can hold 5 to 8 pounds. A large box can hold 8 to
Strike441 [17]

Answer:

Step-by-step explanation: Uhh I have no idea how I got this right because I guessed but

small: 3-5 pounds 0.25+1.2+2.7

Medium: 5-8 pounds 0.25+0.25+1.2+1.2+1.2+1.2

Large 8-10 pounds 0.25+1.2+1.2+1.2+2.7+2.7

Hope I helped this is my first time answering a question have a good day everyone

6 0
1 year ago
A tiling company completes two jobs. The first job has $1200 in labor expenses for 40 hours worked, while the second job has $15
daser333 [38]

Answer:

The correct option is;

1560 = 30(52) + b

Step-by-step explanation:

The cost of the first job = $1,200 in labor and expenses

The number of hours worked in the first job = 40

The second job costs $1,560 in labor and expenses

The number of hours worked in the second job = 52

If the cost of labor per hour = L and the expenses = b, we have;

$1,200 = 40×L + b......................(1)

$1,560 = 52×L + b.......................(2)

Subtracting equation (1) from (2), we have;

52×L + b - (40×L + b) =  52×L - 40×L + b - b = $1,560 - $1,200 = $360

12×L = $360

L = $360/12 = $30/hour

From equation (1), we have;

$1,200 = 40×L + X =  40×$30 + b

Therefore, the equation that can be used to calculate the y-intercept of the linear equation is either;

1200 = 40(30) + X or 1560 = 32(52) + b, which gives the correct option as 1560 = 32(52) + b

6 0
1 year ago
Mrs. Johnson buys 2 chickens. The average weight of the 2 chickens is 4 pounds. One of the chickens is 2 pounds heavier. What is
irakobra [83]
Heaviest chicken is 6 pounds
7 0
1 year ago
Read 2 more answers
How do you this question? Please help.
dangina [55]
N - Nicholes money
M - Marys money

(1) N + M = 4800 ⇒ N = 4800 - M
(2) N - 800 = M - 2/5M ⇒ N - 800 = 3/5M

subtitute (1) to (2)

4800 - M - 800 = 3/5M
4000 - M = 3/5M
-M - 3/5M = -4000
-8/5M = -4000  |multiply both sides by (-5/8)
M = 2500

subtitute the value of M to (1)

N = 4800 - 2500
N = 2300

Answer: Mary had at first $2500. (Nichole: $2300).
5 0
1 year ago
An investor believes that investing in domestic and international stocks will give a difference in the mean rate of return. They
arlik [135]

Answer:

So on this case the 90% confidence interval would be given by -4.137 \leq \mu_1 -\mu_2 \leq 2.087  

For this case since the confidence interval for the difference of means contains the 0 we can conclude that we don't have significant differences at 10% of significance between the two means analyzed.

Step-by-step explanation:

Previous concepts  

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".  

The margin of error is the range of values below and above the sample statistic in a confidence interval.  

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".  

\bar X_1 =2.0233 represent the sample mean 1  

\bar X_2 =3.048 represent the sample mean 2  

n1=15 represent the sample 1 size  

n2=15 represent the sample 2 size  

s_1 =4.893387 population sample deviation for sample 1  

s_2 =5.12399 population sample deviation for sample 2  

\mu_1 -\mu_2 parameter of interest at 0.1 of significance so the confidence would be 0.9 or 90%

We want to test:

H0: \mu_1 = \mu_2

H1: \mu_1 \neq \mu_2

And we can do this using the confidence interval for the difference of means.

Solution to the problem  

The confidence interval for the difference of means is given by the following formula:  

(\bar X_1 -\bar X_2) \pm t_{\alpha/2}\sqrt{\frac{s^2_1}{n_1}+\frac{s^2_2}{n_2}} (1)  

The point of estimate for \mu_1 -\mu_2 is just given by:  

\bar X_1 -\bar X_2 =2.0233-3.048=-1.0247

The degrees of freedom are given by:

df = n_1 +n_2 -2 = 15+15-2=28  

Since the Confidence is 0.90 or 90%, the value of \alpha=0.1 and \alpha/2 =0.05, and we can use excel, a calculator or a table to find the critical value. The excel command would be: "=-T.INV(0.05,28)".And we see that t_{\alpha/2}=\pm 1.701  

Now we have everything in order to replace into formula (1):  

-1.0247-1.701\sqrt{\frac{4.893387^2}{15}+\frac{5.12399^2}{15}}=-4.137  

-1.0247+1.701\sqrt{\frac{4.893387^2}{15}+\frac{5.12399^2}{15}}=2.087  

So on this case the 90% confidence interval would be given by -4.137 \leq \mu_1 -\mu_2 \leq 2.087  

For this case since the confidence interval for the difference of means contains the 0 we can conclude that we don't have significant differences at 10% of significance between the two means analyzed.

4 0
1 year ago
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