Overview Formula Exercise – 1 Exercise – 2 Exercise – 3 Exercise – 4 Exercise – 5 TU Solution

## Exercise – I

## Test of significance of singal mean

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Z- Test Singal Mean Test Setting

Null Hypothessis (H_{0}: mu=mu_{0}) (Given value of (mu) )

There are no significant different between sample mean and population mean

Alternative Hypothessis (H_{1}: mu neq mu_{0}) (Two tail test)

There are significant different between sample mean and population mean (H_{1}: mu>mu_{0}) (Right tail test or one tail test)

Population mean is greater than Sample mean

(H_{1}: mu<mu_{0}) (Left tail test or one tail test)

Population mean is less than Sample mean

Test of statistics,

(Z_{c a l}=frac{bar{X}-mu}{frac{sigma}{sqrt{n}}}=frac{bar{X}-mu}{frac{s}{sqrt{n}}}(hat{sigma}=) s for largr sample ())

Find (Z_{text {tab }}) from table using significance level (1 %, 2 %), or (5 %). If significance level

Is not given then use (5 %)

Coparission and Dissision

Since (left|Z_{c a l}right| leqleft|Z_{t a b}right|) it is not significant and (mathrm{H}_{0}) accepted, (mathrm{H}_{1}) rejected means

The sample mean is same as the Population mean

Since (left|Z_{c a l}right|>left|Z_{t a b}right|) it is significant and (H_{0}) rejected, (H_{1}) accepted means Sample mean is same as Population mean

## Z- Test Double Mean Test Set up

Null Hypothessis (H_{0}: mu_{1}=mu_{2}) (Given value of (mu) )

There are no significant different between two population mean

Alternative Hypothessis (H_{1}: mu_{1} neq mu_{2}) (Two tail test)

There are significant different between two population mean

(H_{1}: mu_{1}>mu_{2}) (Right tail test or one tail test)

First population mean is greater than second one

(H_{1}: mu_{1}<mu_{2}) (Left tail test or one tail test)

First population mean is less than second one

Test of statistics,

(Z_{c a l}=frac{bar{X}_{1}-bar{X}_{2}}{sqrt{frac{sigma_{1}^{2}}{n_{1}}+frac{sigma_{1}^{2}}{n_{2}}}}=frac{bar{X}_{1}-bar{X}_{2}}{sqrt{frac{s_{1}^{2}}{n_{1}}+frac{s_{1}^{2}}{n_{2}}}}left(right.) for large sample (hat{sigma}_{1}=s_{1}) and (left.hat{sigma}_{1}=s_{2}right))

Find (Z_{t a b}) from table using significance level (1 %, 2 %), or (5 %). If significance level is not given then use (5 %)

Coparission and Dissision

Since (left|Z_{c a l}right| leqleft|Z_{t a b}right|)

it is not significant and (mathrm{H}_{0}) accepted, (mathrm{H}_{1}) rejected means There are no significant different between two population mean

Since (left|Z_{c a l}right| leqleft|Z_{t a b}right|)

it is significant and (H_{0}) rejected, (H_{1}) accepted means

There are no significant different between two population mean

## Exercise – II

## Test of significance for single proportions

Step 1

(H_{0}: P=P_{0}). That is, population proportion has some specified value (P_{0}).

(H_{1}: P neq P_{0}) (Two tailed test). That is, population proportion is not equal to (P_{0}).

(mathrm{H}_{1}: mathrm{P}>mathrm{P}_{0}) (Right tailed test). That is, the population proportion is greater than (mathrm{P}_{0}).

(mathrm{H}_{1}: mathrm{P}<mathrm{P}_{0}) (Left tailed test). That is, the population proportion is less than (mathrm{P}_{0}).

It is noted that one should be chosen only one alternative hypothesis depending upon the nature of the problem.

Step – 2

Test Statistics: Under (mathrm{H}_{0 mathrm{Z}}=frac{p-P}{sqrt{frac{P Q}{n}}})

Where, (P=) population proportion of success

(mathrm{p}=) sample proportion of success (=frac{X}{n})

(X=) number of successes

(mathrm{n}=) sample size or no. of trials

SE. ((P)=) Standard error of proportion (=sqrt{frac{P Q}{n}})

Step – 3

Write down the tabulated value of (Z) at (alpha) level of significance according as whether the alternative hypothesis is one tailed test or two tailed tests.

Step-4

Conclusion:

(left|Z_{text {cal }}right| leqleft|Z_{text {tab }}right| rightarrow mathrm{H}_{0}) is accepted (rightarrow) it is not significant

(left|Z_{c a l}right|>left|Z_{t a b}right| rightarrow H_{1}) is accepted (rightarrow) it is not significant

## Test of Significance for the Difference of Two Proportions

Test of Significance for Difference of Two Proportions

Set up the null hypothesis and alternative hypothesis

Step 1

Null hypothesis (mathrm{H}_{0}: mathrm{P}_{1}=mathrm{P}_{2}=mathrm{P}) (say). That is, there is no significant difference between two sample proportions (p_{1}) and (p_{2}).

Alternative hypothesis (H_{1}: P_{1} neq P_{1}) (Two tailed test). That is, there is significant difference between two sample proportions (p_{1}) and (p_{2}).

(H_{1}: P_{1}>P_{1}) (Right tailed test). That is, one group population proportion is greater than other group population proportion.

(H_{1}: P_{1}<P_{2}) (Left tailed test). That is, one group population proportion is less than other group population proportion.

Step 2

Test statistic: Under (mathrm{H}_{0}: mathrm{P}_{1}=mathrm{P}_{2}=hat{P}_{text {, then test statistic is }}^{text {, }}) is

(Z_{text {cal }}=frac{P_{1}-P_{2}}{sqrt{P Qleft(frac{1}{n_{1}}+frac{1}{n_{2}}right)}})

where (P=) the common population proportion under (mathrm{H}_{0}) is unknown and we use its unbiased estimate provided by both samples taken together which is given by (hat{P}=frac{X_{1}+X_{2}}{n_{1}+n_{2}}=frac{n_{1} p_{1}+n_{2} p_{2}}{n_{2}+n_{2}})

And (hat{Q}=1-hat{P})

If (P) unknown then test statistics is

(Z_{mathrm{cal}}=frac{p_{1}-p_{2}}{sqrt{hat{P} hat{Q}left(frac{1}{n_{1}}+frac{1}{n_{2}}right)}})

Step 3

Obtain tabulated (critical) value of Z at (alpha) level of significance for appropriate alternative hypothesis.

The most commonly used is (alpha=5 %).

Step 4

Decision: Make a decision by comparing the calculated value of (Z) with critical value of (Z).

(left|Z_{text {cal }}right| leqleft|Z_{t a b}right| rightarrow H_{0}) is accepted (rightarrow) it is not significant

(left|Z_{text {cal }}right|>left|Z_{text {tab }}right| rightarrow H_{1}) is accepted (rightarrow) it is not significant.