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≤ |