An engineer conducted an experiment to compare two metals: iron and copper, as bonding agents for an alloy material. Components of the alloy were bonded using the metals as bonding agents, and the pressures required to break the bonds were measured. The data for the breaking pressures are given in the following table:
| S. No | Breaking Pressure | |
| Iron | Copper | |
| 1 | 72.7 | 73 |
| 2 | 69.6 | 67.2 |
| 3 | 83.4 | 75.3 |
| 4 | 78.9 | 61.4 |
| 5 | 75 | 74 |
| 6 | 71.6 | 69.5 |
| 7 | 85.7 | 69.8 |
| 8 | 73.5 | 73.8 |
| 9 | 70.4 | 68 |
| 10 | 84.2 | 76.1 |
If the breaking pressures for both iron and copper are normally distributed, are the variances of the distributions of the breaking pressure of iron and copper equal at 5 % level of significance?
We are asked to test whether the variances of breaking pressures for iron and copper are equal at a 5% significance level. This is an F-test for equality of variances. Let’s solve step by step manually. Step 1: Collect data Iron: 72.7, 69.6, 83.4, 78.9, 75, 71.6, 85.7, 73.5, 70.4, 84.2 Copper: 73, 67.2, 75.3, 61.4, 74, 69.5, 69.8, 73.8, 68, 76.1 Step 2: Compute means Mean of iron ($\bar{X}_1$): $$ \bar{X}_1 = \frac{72.7+69.6+83.4+78.9+75+71.6+85.7+73.5+70.4+84.2}{10} $$ Step by step addition: * 72.7 + 69.6 = 142.3 * 142.3 + 83.4 = 225.7 * 225.7 + 78.9 = 304.6 * 304.6 + 75 = 379.6 * 379.6 + 71.6 = 451.2 * 451.2 + 85.7 = 536.9 * 536.9 + 73.5 = 610.4 * 610.4 + 70.4 = 680.8 ________ ______ ____ _______ _________ _____ ______ ____.
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A baby-sister has 6 children under her supervision. The age of each child is as follows:
| Child | Age (in years |
| Sonu | 2 |
| Lavnik | 4 |
| Chiya | 3 |
| Amam | 3 |
| Avishi | 4 |
| Ridhi | 5 |
| Sidhi | 3 |
Complete the following table, one is done for you:
| S. No. | Test For | Name of the Test | Test Statistic |
| 1 | Population mean when population variance is known and population is normal | Z-test | |
| 2 | Population mean when population variance is unknown and population is normal | ||
| 3 | Difference of two population means when samples are paired, and population of differences follows normal distribution. | ||
| 4 | Difference of two population means when samples are independent, and population of differences follows normal distribution. | ||
| 5 | Population variance when the population is normal distributed | ||
| 6 | Population variance when the population is not normal distributed |
Describe the various forms of the sampling distribution single mean.
An engineer conducted an experiment to compare two metals: iron and copper, as bonding agents for an alloy material. Components of the alloy were bonded using the metals as bonding agents, and the pressures required to break the bonds were measured. The data for the breaking pressures are given in the following table:
| S. No | Breaking Pressure | |
| Iron | Copper | |
| 1 | 72.7 | 73 |
| 2 | 69.6 | 67.2 |
| 3 | 83.4 | 75.3 |
| 4 | 78.9 | 61.4 |
| 5 | 75 | 74 |
| 6 | 71.6 | 69.5 |
| 7 | 85.7 | 69.8 |
| 8 | 73.5 | 73.8 |
| 9 | 70.4 | 68 |
| 10 | 84.2 | 76.1 |
State whether the following statements are True or False. Give reasons in support of your answer:
(i) is a random sample of size 5 taken from an Exponential distribution, then estimator T₁ is more efficient than T2.
(ii) are two estimators of the parameter such that Var(T1) = 1/n and Var(T2) = n then T₁ is more efficient than T2.
(iii) A 95% confidence interval is smaller than a 99% confidence interval.
(iv) If the probability density function of a random variable X follows F-distribution is
then degrees of freedom of the distribution will be (2,2).
(v) A patient suffering from fever reaches to a doctor and suppose the doctor formulates the hypotheses as
Ho: The patient is a chikungunya patient
H₁: The patient is not a chikungunya patient
If the doctor rejects Ho when the patient is actually a chikunguniya patient, then the doctor commits type II error.
The department of transportation has mandated that the average speed of cars on interstate highways be no more than 70 km per hour in order. To check whether the people
followed it or not, a researcher took a random sample of 186 cars and found that the average speed was 72 km per hour with a standard deviation 0.6 km per hour.
(i) State null and alternative hypotheses.
(ii) Name the test which is suitable in this situation and why?
(iii) Calculate the value of test statistic and critical value.
(iv)Draw the conclusion on the basis of the applied test.