Testing of hypothesis

T-Test for Difference Between Two Sample Means

Hypothesis Testing: Type I and Type II Errors

Define Type I and Type II error in testing of hypothesis. A psychologist wishes to verify that a certain drug increases the reaction time to given stimulus. The following reaction times (in tenth of seconds) were recorded before and after injection of the drug for each of four subjects. Test at 5% level of significance to determine whether the drug significantly increases the reaction time.

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The mean drying time of a brand of spray paint is known to be 122 seconds. The research division of the company that produces this paint contemplates that adding a new chemical ingredient to the paint accelerate the drying process. To investigate this conjecture, the paint with the chemical additions is sprayed on 50 surfaces and the drying time is recorded. The mean and standard deviation of drying time computed from these recorded are found as 116 seconds and 16.8 seconds respectively. Does these data provide strong evidence that the mean drying time is reduced by the addition of the new chemical? Use 5% level of significance. Also find p-value. [5]

Hypothesis Testing for Average Age of CSIT Students

Step 1: Set up Hypotheses

• Null Hypothesis (H₀): $\mu = 22$ (The average age of enrolling students is 22 years)
• Alternative Hypothesis (H₁): $\mu < 22$ (The average age is less than 22 years)

This is a left-tailed test at 5% level of significance.

Step 2: Sample Data

| 20 | 19 | 22 | 23 | 19 | 20 | 20 | 21 | 22 | 20 | 19 | 20 |

Sample size: $n = 12$

Step 3: Calculate Sample Statistics

Sample mean: $$\bar{x} = \frac{20 + 19 + 22 + 23 + 19 + 20 + 20 + 21 + 22 + 20 + 19 + 20}{12} = \frac{245}{12} = 20.417$$

Sample standard deviation: $$s = \sqrt{\frac{\sum(x_i - \bar{x})^2}{n-1}} = \sqrt{\frac{26.917}{11}} = \sqrt{2.447} = 1.564$$

Step 4: Test Statistic

Since the population is normally distributed and population standard deviation is unknown, we use the t-test:

$$t = \frac{\bar{x} - \mu_0}{s/\sqrt{n}} = \frac{20.417 - 22}{1.564/\sqrt{12}} = \frac{-1.583}{0.451} = -3.511$$

Step 5: Critical Value

For a left-tailed test with $\alpha = 0.05$ and $df = n - 1 = 11$: $$t_{0.05, 11} = -1.796$$

Step 6: Decision

Since $t = -3.511 < -1.796$, we reject the null hypothesis.

Conclusion: At the 5% level of significance, there is sufficient evidence to support the researcher's doubt that the average age of CSIT enrolling students is less than 22 years. [5]

A dealer of a DELL company located at New Road claimed that the average lifetime of a multimedia projector produced by Dell Company is greater than 60,000 hours with standard deviation of 6000 hours. In order to test his claim, sample of 100 DELL projectors are taken and the average life time was monitored and it was found to be 55,000 hours. Test the claim of the dealer at 5% level of significance. [5]

Discuss the concept of level of significance in hypothesis testing. A manufacturer of laptop provides a particular model in one of three colors. Of the first 100 laptops sold, it is noted that 80 were the first color. Can you conclude that more than two third of all the customers have a preference for the first color? Use 5% level of significance. [5]

Define type I and type II error in testing of hypothesis. It is claimed that Samsung and Huawei mobiles are equally popular in Kathmandu. A random sample of 600 people from Kathmandu showed 350 have Samsung mobile. Test the claim at 5% level of significance. [5]

It is claimed that Samsung and Redmi mobiles are equally popular in Kathmandu. A random sample of 500 people from Kathmandu showed 300 have Samsung mobile. Test the claim at 5% level of significance. [5]

The following are the details of working hours in the classroom per week of male and female faculty working in the area of Computer Science and Information Technology at Tribhuvan University. Apply independent t-test to examine the average working hour in the classroom per week is significantly different between male and female faculty, at 1% level of significance. State also null and alternative hypotheses appropriately.

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In location 1, there are 250 corona-positive cases out of 460 persons, and in location 2, 250 positive cases were reported out of 650 persons. Can it be concluded that the proportion of corona-positive cases is higher in location 1 compared to location 2? Test at a 10% level of significance. [5]

What do you mean by hypothesis? Describe null and alternative hypothesis. A company claims that its light bulbs are superior to those of the competitor on the basis of study which showed that a sample of 40 of its bulbs had an average life time 628 hours of continuous use with a standard deviation of 27 hours. While sample of 30 bulbs made by the competitor had an average life time 619 hours of continuous use with a standard deviation of 25 hours. Test at 5% level of significance, whether this claim is justified.[10]

Independent t-test and Confidence Interval Analysis

Based on the following information, perform the following:

a. Test whether two means are significantly different (α = 5%) using independent t-test.

b. Compare 95% confidence interval estimation for the difference of means.

c. Show the linkage between testing of hypothesis and confidence interval estimation in this problem.

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