Mathematics

Solved Prob 1

An insurance company is entering in a new area, which is demographically very similar to other areas where it sells insurance. The average based on these areas is $20450. They are, however, concerned that the claims in this new area may actually be higher. They randomly select 50 claims, and calculate a sample mean of $21050. Assuming that the standard deviation of claims is $4500, and set α = :05, test to see if the insurance company’s concern is valid?
Answer:
Here you need to first decide what is the null hypothesis and what is the alternative hypothesis.
The question we are answering is whether the insurance company should be concerned?
The company would be concerened if the average claims have gone up.
Thus the alternative hypothesis is
HA μ>20450
The null hypothesis, thus is μ=1800. You may also write it as The null hypothesis, thus is μ≤1800
Since the population standard deviation is known (in the problem it is stated ‘assume it is $500’), we would use Normal Distribution.
Thus calculated value of the statistics is
z =(21550-20450)/(4500/SQRT(50))
1.7284832429
The next question is ‘what is the decision rule?’
Should the cutoff point be 1.96 or 1.645?
Remember, this is a one sided test, i.e., we reject only if the value is in the rejection range.
Thus, using α=.05, the cutoff point would be 1.645, since probability (area) in the rejection region is .05.
If we were using two sided test, we would not reject the null hypothesis.
Another way of answering this would be to calculate the p value.
In this case, it would be prob
=1-NORM.S.DIST(B20,1)
0.0419508114
Since this is less than .05 we would reject the null hypothesis.
Third way of calculating this would be to calculate this directly, without calculating the z.
=1-NORM.DIST(21550,20450,4500/SQRT(50),1)
0.0419508114
Again, since this is less than .05, we would reject the null hypothesis.

Prob 2

Problem 2 When you moved, your friend told you that the average weekly grocery bill in your town is $150. You believe that it is higher. You took a sample of 25 individuals and asked them for their grocery bills and found that the average is $160.3 and the standard deviation for the sample is $29.5. Set up your null and alternative hypothesis and do the analysis using α=.05.
Again, here your alternative is
HA μ>150
The null, thus, is
H0 μ=150
Since we do not know the population standard deviation, we would use sample standard deviation and, instead of the normal distribution, use the t distribution.
Calculated t =(160.3-150)/(29.5/SQRT(25))
1.7457627119
How do you calcuate the tabulated t?
d.f. = 25 – 1, i.e., 24
1.7108820799
Since the calculated value of 1.745 is greater than tabulated value 1.710882, we would reject the null nypothesis.
Again, as an alternative, you could have calculate the tail probability as follows:
0.0468216589
Since this value is less than .05, we would reject the null hypothesis.

Prob 3

Problem 3 Your average weekly grocery bill at the store you usually shop is $160. A new grocery store has opened near you. You want to know if your grocery bill would be any different if you shop there. You took a sample of 55 individuals and asked them for their grocery bills and found that the average is $149.5 and the standard deviation for the sample is $25.5. Set up your null and alternative hypothesis and do the analysis using α=.05.
Since you are interesting in knowing whether it any different, your alternative hypothesis is
HA μ≠ 160
The null, thus, is
H0 μ= 160
Also, n= 55 s= 25.5
Since we do not know the population standard deviation, we would use sample standard deviation and, instead of the normal distribution, use the t distribution.
Calculated t =(149.5-d11)/(f12/SQRT(d12))
-3.0537287888
How do you calcuate the tabulated t?
d.f.= n-1= 54
Tabulated t 2.0048792882
Thus, since caculated t-=3.054<tabulated value of 2.0049, you would reject the null hypothesis.
You can calculate the tail probability as probability to the left of the calculated t value and multiply by 2.
Tail Prob. = 0.002
p value = 0.004
Since this value is less than .05, we would reject the null hypothesis.

Prob 4

Problem 5 You started a wedding planning business. Based on expenses of 25 weddings, you estimated costs of a small wedding to be $29,500. To provide a range, you decided to provide a 95% confidence interval and searched for an estimate of standard deviation of wedding expenses and believe that it is $8000. Prepare a 95% confidence interval for the averge cost of wedding.
If the confidence level is 95%, then error probability, i.e., α=.05.
How do you calcuate the tabulated z?
This is something you had calculate in week2.
You do inverse of the normal distribution for a probability of .975. I.e., .025 is on the left and .025 is not the right. If you exclude, .025 on the right, you are left with .975.
Thus, required z value is =NORM.S.INV(0.975)
1.9599639845
Using this value of zα/2, the lower confidence limit would be
= =29500 – 1.96*(8000/SQRT(25))
26364
Similarly, the Upper Confidence Interval would be
= =29500 + 1.96*(8000/SQRT(25))
32636
The confidence interval, thus, is as follows:
26364 ≤μ≤ 32636

Prob 5

Problem 5 A new grocery store has opened near you. You want to be 95 % sure to find a range within which your total weekly bill would lie. You took a sample of 55 individuals and asked them for their grocery bills and found that the average is $149.5 and the standard deviation for the sample is $25.5. Prepare a 95% confidence interval for the true weekly expenditure.
If the confidence level is 95%, then error probability, i.e., α=.05.
How do you calcuate the tabulated t?
d.f.= n-1= 54
Tabulated t 2.0048792882
Using this value of t, the lower confidence limit would be
= =149.5 – 2.0048*(25.5/SQRT(55))
142.6066569902
Similarly, the Upper Confidence Interval would be
= =149.5 + 2.0048*(25.5/SQRT(55))
156.3933430098
The confidence interval, thus, is as follows:
142.6066569902 ≤μ≤

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