# an insurance agent has claimed that the average age of policy holders who insure through him is less than the average for all the agents, which is 35 years. a random sample of 40 policy holders who insured through him gave an average of 32 years with the standard error of 2 years. using α at 5% level of significance, ascertain whether the insurance agents claim is justified? also find p-value.

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get an insurance agent has claimed that the average age of policy holders who insure through him is less than the average for all the agents, which is 35 years. a random sample of 40 policy holders who insured through him gave an average of 32 years with the standard error of 2 years. using α at 5% level of significance, ascertain whether the insurance agents claim is justified? also find p-value. from screen.

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18. An insurance agent has claimed that the average age of policy-holders who insure through him is-less than the average for all agents, which is 30-5 years. A random sample of 100 policy-holders who had insured through him gave the following age distribution : Age last birthday 16-20 21-25 26-30 31-35 36- 40 No. of persons 12 22 20 30 16 Calculate the arithmetic mean and standard deviation of this distribution and use these values to test his claim at the 5% level of significance. You are given that Z (1-645) = (0-95.

## 18. An insurance agent has claimed that the average age of policy-holders who insure through him is-less than the average for all agents, which is 30-5 years. A random sample of 100 policy-holders who had insured through him gave the following age distribution : Age last birthday 16-20 21-25 26-30 31-35 36- 40 No. of persons 12 22 20 30 16 Calculate the arithmetic mean and standard deviation of this distribution and use these values to test his claim at the 5% level of significance. You are given that Z (1-645) = (0-95.

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Transcribed Image Text:18. An insurance agent has claimed that the average age of policy-holders who insure through him is less than the average for all agents, which is 30-5 years. A random sample of 100 policy-holders who had insured through him gave the following age distribution : Age last birthday 16-20 21-25 26-30 31-35 36-40 No. of persons 12 22 20 30 16 : Calculate the arithmetic mean and standard deviation of this distribution and use these values to test his claim at the 5% level of significance. You are given that Z (1-645) = (0-95.

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**Question:** 1. An Insurance Agent Has Claimed That The Average Age Of Policy Holders Who Insure Through Him Is Less Than The Average For All The Agents, Which Is 35 Years. A Random Sample Of 40 Policy Holders Who Insured Through Him Gave An Average Of 32 Years With The Standard Error Of 2 Years. Using Α At 5% Level Of Significance, Ascertain Whether The Insurance Agents

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## Expert Answer

1st step All steps Final answer Step 1/4 Given : Sample mean X―=32 years Sample SE sn=2 years Sample size n=40

Level of significance

α=5%=0.05

Here we have to test the claim that the average age of policy holders who insure through insurance agent is less than 35 i.e. the average for all the agents.

To test the above claim the null and alternative hypothesis are,

H0:μ≥35

The average age of policy holders who insure through insurance agent is greater than equal to 35 .

vs H1:μ<35

The average age of policy holders who insure through insurance agent is less than 35 i.e. the average for all the agents.

Explanation for step 1

The null hypothesis is hypothesis of no difference and always contains an equality sign. The alternative hypothesis is opposite to the null hypothesis.

The standard error is given as,

SE=sn

where s is sample standard deviation.

View the full answer

Step 2/4 Step 3/4 Step 4/4 Final answer

Transcribed image text:

1. An insurance agent has claimed that the average age of policy holders who insure through him is less than the average for all the agents, which is 35 years. A random sample of 40 policy holders who insured through him gave an average of 32 years with the standard error of 2 years. Using α at 5% level of significance, ascertain whether the insurance agents claim is justified? Also find P-value.

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## Testing of Hypotheses, Quantitative Techniques (MB

Question Paper Solutions of Testing of Hypotheses, Quantitative Techniques (MB - 106), 1st Semester, Master of Business Administration, Maulana Abul Kalam Azad University of Technology

**Maulana Abul Kalam Azad University of Technology**

### MB - 106 - Quantitative Techniques

### Testing of Hypotheses

Group B

**Repeated:**2008

Marks: 15

**1.**A machine produced 20 defective items in a batch of 450. After overhauling it produced 10 defective items in a batch of 300. Has the machine overhauled? At a 5 % level of significance z = 1.64 for one-tailed test and z = 1.96 for two-tailed test.

**Repeated:**2009

Marks: 15

**2.**Two samples are drawn from two normal populations. From the following data, test whether the two samples have the same variance at the 5% level.

**Sample I**60 65 71 74 76 82 85 87

**Sample II**61 66 67 85 78 63 85 86 88 91

(Given the tabulated value of at 5% level with 9 and 7 degrees of freedom is 3.68).

**Repeated:**2009

Marks: 15

**3.**Two groups X and Y consist of 100 people each of who have a disease. A medicine is given to group X but not to Y. people respectively recover It is found that in groups X and Y, 75 and 65 from medicine helps to cure the disease. Test. the hypothesis that the disease. (Use a 5% level where the critical value of Z is 1.96).

**Repeated:**2010,2013

Marks: 15

**4.**A manufacturer's ball-point refills have a mean life of 400ages with an S.D of 2 Pages. A purchasing agent selects a sample of 100 pens and put them for the test. The mean writing life for the sample was found to be 39 pages. Should the purchasing agent reject the manufacturer's claim at a 5% level of significance?

**Repeated:**2010

Marks: 15

**5.**A group of 5 patients, treated with medicine A weigh 42 kg, 39 kg, 48 kg, 60 kg, 41 kg. The second group of 7 patients from the same hospital treated with medicine. B weight 38 kg, 42 kg, 56 kg, 64 kg, 68 kg, 69 kg, 62 kg. Do you agree with the claim that medicine B increases the weight significantly? (The value of at a 5% level of significance for 10 degrees of freedom is 2.2281).

**Repeated:**2010

Marks: 15

**6.**In order to test whether a coin is perfect, it is tossed 5 times. The null hypothesis Of perfectness is rejected if and times only if more than 4 heads is obtained. Obtain the

i) probability of type I error and

ii) probability of type Il error when the corresponding probability of getting a head is 0.2.

**Repeated:**2012

Marks: 15

**7.**An insurance agent has claimed that the average age of policyholders who insure through him is less than the average for all agents, which is 30.5 years. A random sample of 100 policyholders who had insured through him gave the following age distribution:

**Age (years)**

**16 - 20**

**21 - 25**

**26 - 30**

**31 - 35**

**36 - 40**

**Total**

**No. of persons**12 22 20 30 16 100

Test his claim at a 5% level.

**Repeated:**2014

Marks: 15

**8.**Two random samples of size 9 and 13 are drawn from a population and their sample S.D. are as follows:

**Sample**

**Size**

**S.D.**

1 9 2.1 8 13 1.8

Can the samples be regarded as drawn from a normal population with the same S.D. Given a 5% value of F for (8.12) d.f. is 2.85.

**Repeated:**2016

Marks: 15

**9.**A coin has tossed 4000 times, which turns up head 219 times. Do the data justify the hypothesis of an unbiased coin?

Group B

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