7/5+2/5x=31/15+7/3x+1/3
1 answer:
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Answer:0.1446
Step-by-step explanation:
Let p be the population proportion of adults say that cloning of humans should be allowed.
As per given , the appropriate set of hypothesis would be :-
Since is left-tailed and sample size is large(n=1012) , so we perform left-tailed z-test.
Given : The test statistic for this test is -1.06.
P-value for left tailed test,
P(z<-1.06)=1-P(z<1.06) [∵P(Z<-z)=1-P(Z<z]
= (Using z-value table.)
Hence, the p-value for this test statistic= 0.1446
Answer:
The 99% confidence interval estimate of the percentage of girls born is (74.37%, 85.63%).
Usually, 50% of the babies are girls. This confidence interval gives values considerably higher than that, so the method to increase the probability of conceiving a girl appears to be very effective.
Step-by-step explanation:
Confidence Interval for the proportion:
In a sample with a number n of people surveyed with a probability of a success of , and a confidence level of
, we have the following confidence interval of proportions.
In which
z is the zscore that has a pvalue of .
For this problem, we have that:
99% confidence level
So , z is the value of Z that has a pvalue of
, so
.
The lower limit of this interval is:
The upper limit of this interval is:
For the percentage:
Multiplying the proportions by 100.
The 99% confidence interval estimate of the percentage of girls born is (74.37%, 85.63%).
Usually, 50% of the babies are girls. This confidence interval gives values considerably higher than that, so the method to increase the probability of conceiving a girl appears to be very effective.