# 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≤ |