General Procedure for Test of Hypotheses for Population Proportion: Procedure Steps, Example Solved Problems

**TEST OF HYPOTHESES FOR POPULATION PROPORTION**

**Step 1 : **Let** ***P*** **denote the proportion of the population
possessing the qualitative characteristic** **(attribute) under study. If *p _{0}*
is an admissible value of

(i) *H _{1}: P ≠ p_{0}* (ii)

**Step 2 : **Let** ***p*** **be proportion of the sample
observations possessing the attribute, where** ***n*** **is** **large,
*np* > 5 and *n(*1 *– p)* > 5.

**Step 3 : **Specify the level of significance,** ***α*.

**Step 4 : **Consider the test statistic *Z*** **** **under *H*_{0}. Here, *Q* = 1 – *P*.

The approximate sampling distribution of the test
statistic under *H*_{0} is the *N(0,1)* distribution.

**Step 5 : **Calculate the value of *Z* under *H*_{0} for the
given data as

**Step 6 :** Choose the critical value, *z*_{e}, corresponding to
*α* and *H*_{1} from the following table

**Step 7 :** Make decision on *H*_{0} choosing the suitable
rejection rule from the following table corresponding to *H*_{1}.

A survey was conducted among the citizens of a city to study their
preference towards consumption of tea and coffee. Among 1000 randomly selected
persons, it is found that 560 are tea-drinkers and the remaining are
coffee-drinkers. Can we conclude at 1% level of significance from this
information that both tea and coffee are equally preferred among the citizens
in the city?

**Step 1 : **Let** ***P*** **denote the proportion of people in the
city who preferred to consume tea.

Then, the null and the alternative hypotheses are

**Null hypothesis: ***H*** **_{0}** **:** ***P*** **=** **0.5

*i.e*., it is significant that both tea and coffee are preferred
equally in the city.

**Alternative hypothesis: ***H*** **_{1}** **:** ***P*** **≠** **0.5

*i.e*., preference of tea and coffee are not significantly equal. It is
a two-sided alternative* *hypothesis.

**Step 2 : ****Data**

The given sample information are

Sample size (*n*) = 1000. Hence, it is a large sample.

No. of tea-drinkers = 560

Sample proportion (*p*) = 560/1000 = 0.56

**Step 3 : ****Level of significance**

α= 1%

**Step 4 : ****Test statistic**

Since *n* is large, *np* = 560 > 5 and *n*(1 – *p*)
= 440 > 5, the test statistic under the null hypothesis, is Z = .

Its sampling distribution under *H*_{0} is the *N(0,1)*
distribution.

**Step 5 : ****Calculation of Test Statistic**

The value of *Z* can be calculated for the sample information
from

Thus, *z*_{0} =
3.79

**Step 6 : ****Critical value**

Since *H*_{1} is a two-sided alternative hypothesis,
the critical value at 1% level of significance is *z _{α/}*

**Step 7 : ****Decision**

Since *H*_{1} is a two- sided alternative, elements
of the critical region are determined by the rejection rule |*z*_{0}|
≥ *z* * _{e}*. Thus it is a two-tailed test. Since |

Tags : Procedure Steps, Example Solved Problems | Statistics , 12th Statistics : Chapter 1 : Tests of Significance - Basic Concepts and Large Sample Tests

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12th Statistics : Chapter 1 : Tests of Significance - Basic Concepts and Large Sample Tests : Test of Hypotheses for Population Proportion | Procedure Steps, Example Solved Problems | Statistics

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