Hypothesis Problems

Q1. A sample of 400 male students is found to have a mean height 67.47 inches. Can it be reasonably regarded as a sample from a large population with mean height 67.39 inches and standard deviation1.30 inches? Test at 5% level of significance.

Solution: Taking the null hypothesis that the mean height of the population is equal to 67.39 inches, we can write:

and the given information as X = 67.47”, σp = 1.30" , n = 400. Assuming the population to be normal, we can work out the test statistic z as under:

As Ha is two-sided in the given question, we shall be applying a two-tailed test for determining the

rejection regions at 5% level of significance which comes to as under, using normal curve area table:

R: | z | > 1.96

The observed value of z is 1.231 which is in the acceptance region since R : | z | > 1.96 and thus  H0 is accepted. We may conclude that the given sample (with mean height = 67.47") can be regarded

Q2 Suppose we are interested in a population of 20 industrial units of the same size, all of which are experiencing excessive labor problems. The past records show that the mean of the distribution of annual turnover is 320 employees, with a standard deviation of 75 employees. A sample of 5 of these industrial units is taken at random which gives a mean of annual turnover as 300 employees. Is the sample mean consistent with the population mean? Test at 5% level.

Solution: Taking the null hypothesis that the population mean is 320 employees, we can write:

                                                                

and the given information as under:

X = 300 employees, σp =75 employees

n = 5; N = 20

Assuming the population to be normal, we can work out the test statistic z as under:

As Ha is two-sided in the given question, we shall apply a two-tailed test for determining the

rejection regions at 5% level of significance which comes to as under, using normal curve area table:

R: | z | > 1.96

The observed value of z is –0.67 which is in the acceptance region since R: | z | > 1.96 and thus, H0 is accepted and we may conclude that the sample mean is consistent with population mean i.e., the population mean 320 is supported by sample results.

Q3 The mean of a certain production process is known to be 50 with a standard deviation of 2.5. The production manager may welcome any change is mean value towards higher side but would like to safeguard against decreasing values of mean. He takes a sample of 12 items that gives a mean value of 48.5. What inference should the manager take for the production process on the basis of sample results? Use 5 per cent level of significance for the purpose.

Solution: Taking the mean value of the population to be 50, we may write:

* Being a case of finite population.

and the given information as X = 48.5, σp = 2.5 and n = 12. Assuming the population to be normal, we can work out the test statistic z as under:

As Ha is one-sided in the given question, we shall determine the rejection region applying onetailedtest (in the left tail because Ha is of less than type) at 5 per cent level of significance and it

comes to as under, using normal curve area table:

R : z < – 1.645

The observed value of z is – 2.0784 which is in the rejection region and thus, H0 is rejected at 5 per cent level of significance. We can conclude that the production process is showing mean which is significantly less than the population mean and this calls for some corrective action concerning the said process.

Q4 The specimen of copper wires drawn form a large lot have the following breaking strength (in kg.weight):

578, 572, 570, 568, 572, 578, 570, 572, 596, 544

Test (using Student’s t-statistic) whether the mean breaking strength of the lot may be taken to be578 kg. weight (Test at 5 per cent level of significance). Verify the inference so drawn by using Sandler’s A-statistic as well.

Solution: Taking the null hypothesis that the population mean is equal to hypothesized mean of 578 kg., we can write:

As the sample size is mall (since n = 10) and the population standard deviation is not known, we shall use t-test assuming normal population and shall work out the test statistic t as under:

To find X and ss we make the following computations:

Degree of freedom = (n – 1) = (10 – 1) = 9

As Ha is two-sided, we shall determine the rejection region applying two-tailed test at 5 per cent level of significance, and it comes to as under, using table of t-distribution* for 9 d.f.:

R : | t | > 2.262

As the observed value of t (i.e., – 1.488) is in the acceptance region, we accept H0 at 5 percent level and conclude that the mean breaking strength of copper wires lot may be taken as 578 kg weight.

Q5 Raju Restaurant near the railway station at Falna has been having average sales of 500 tea cups per day. Because of the development of bus stand nearby, it expects to increase its sales. During the first12 days after the start of the bus stand, the daily sales were as under:

550, 570, 490, 615, 505, 580, 570, 460, 600, 580, 530, 526

On the basis of this sample information, can one conclude that Raju Restaurant’s sales have increased? Use 5 per cent level of significance.

Solution: Taking the null hypothesis that sales average 500 tea cups per day and they have not increased unless proved, we can write:

As the sample size is small and the population standard deviation is not known, we shall use t-test assuming normal population and shall work out the test statistic t as:

Degree of freedom = n – 1 = 12 – 1 = 11

As Ha is one-sided, we shall determine the rejection region applying one-tailed test (in the right tail because Ha is of more than type) at 5 per cent level of significance and it comes to as under, using

table of t-distribution for 11 degrees of freedom:

R : t > 1.796

The observed value of t is 3.558 which is in the rejection region and thus H0 is rejected at 5 percent level of significance and we can conclude that the sample data indicate that Raju restaurant’s sales have increased.

Table to be referred for problems