In the 1980s, it was generally believed that congenital abnormalities affected about 5% of the nation's children. Some people believe that the increase in the number of chemicals in the environment has led to an increase in the incidence of abnormalities. A recent study examined 384 children and found that 46 of them showed signs of an abnormality. Is this strong evidence that the risk has increased? a) Write appropriate hypotheses. b) Check the necessary assumption s and conditions. c) Perform the mechanic s of the test. What is the P-value? d) Explain carefully what the P-value mean s in context. e) What 's your conclusion? f) Do environmental chemicals cause congenital abnormalities?

A) Null Hypothesis; H0:p = 0.05

Alternative Hypothesis; HA:p > 0.05

B) All conditions are met.

C) P-value = 0.0001

D) the value means that there is a less than 0.01% chance of obtaining 46 children or more with signs of an abnormality among 384 children when 5% of the children have abnormality.

E) We conclude that there is sufficient evidence to support the claim that the risk has increased.

F) No. The environmental chemicals don't cause congenital abnormalities because there could be other variables too.

Step-by-step explanation:

We are given;

p = 5% = 0.05

n = 384

x = 46

We will assume a confidence interval of 95%

A) From the question, the claim is that the population proportion has increased from 5%.

Thus, the hypotheses is stated as;

Null Hypothesis; H0:p = 0.05

Alternative Hypothesis; HA:p > 0.05

B) np = 0.05 x 384 = 19.2

n(1 - p) = 384(1 - 0.05) = 364.8

For the conditions;

Using 10% conditon;

np and n(1 - p) are more than 10 thus the condition is satisfied.

Using success failure condition :it's similar to the 10% condition and thus is met too

Using random sample condition, we assume that the sample is a random sample

Thus; All conditions are met.

C) Let's first calculate sample proportion from;

p^ = x/n

p^ = 46/384

p^ = 0.1198

The formula for the z-value is;

z = (p^ - p)/√(p(1 - p)/n)

z = (0.1198 - 0.05)/√(0.05(1 - 0.05)/384)

z = 0.0698/0.011122

z ≈ 6.28

The P-value will be;

P = P(Z > 6.28)

For ease of reading on the z table, this can also be written as;

P = P(Z > 6.28) = 1 - P(Z < 6.28)

From z-tables P(Z < 6.28) is 0.9999

Thus;

P-value = 1 - 0.9999 = 0.0001 = 0.01%

D) we have a P-value of 0.01%. This means that there is a less than 0.01% chance of obtaining 46 children or more with signs of an abnormality among 384 children when 5% of the children have abnormality.

E) Our P-value of 0.0001 is less than our significance level of 0.05.

Thus,we reject the null hypothesis and conclude that there is sufficient evidence to support the claim that the risk has increased.

F) No. The environmental chemicals don't cause congenital abnormalities because there could be other variables too.

two data sets with the same median will not have the same varaibility

step-by-step explanation: