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.