A 0.01 significance level is used for a hypothesis test of the claim that when parents a particular method of gender selection, the proportion of baby girls is different from 0.5. Assume that sample data consists of 45 girls in 81 births, so the sample statistic of results in a z score that is 1 standard deviation above O. Complete parts (a) through (h) below. a. Identify the null hypothesis and the alternative hypothesis. Choose the correct answer below. b. What is the value of alpha ? (Type an integer or a decimal.) c. What is the sampling distribution of the sample statistic? d. Is the test two-tailed, left-tailed, or right-tailed? e. What is the value of the test statistic? The test statistic is E (Type an integer or a decimal.) f. What is the P-value? The P-value is (Round to four decimal places as needed.) g. What are the critical value(s)? The critical value(s) is/are (Round to three decimal places as needed. Use a comma to separate answers as needed.) h. What is the area of the critical region? The area is r (Round to two decimal places as needed.)
Z-Test For Proportion
Set Up Hypothesis
Null, H0:P=0.5
Alternate, H1: P!=0.5
Test Statistic
No. Of Success chances Observed (x)=45
Number of objects in a sample provided(n)=81
No. Of Success Rate ( P )= x/n = 0.5556
Success Probability ( Po )=0.5
Failure Probability ( Qo) = 0.5
we use Test Statistic (Z) for Single Proportion = P-Po/Sqrt(PoQo/n)
Zo=0.55556-0.5/(Sqrt(0.25)/81)
Zo =1
| Zo | =1
Critical Value
The Value of |Z | at LOS 0.01% is 2.58
We got |Zo| =1 & | Z | =2.58
Make Decision
Hence Value of |Zo | < | Z | and Here we Do not Reject Ho
P-Value: Two Tailed ( double the one tail ) - Ha : ( P != 1 ) = 0.31731
Hence Value of P0.01 < 0.3173,Here We Do not Reject Ho
ANS) H0:P=0.5 , P!=0.5
ALPHA= 0.01,
NORMAL DISTRIBUTION
TWO TAILED
Zo =1
-2.58,+2.58
-2.58 < Z TAB < 2.58
Homework Sourse
