In the 1980s it was generally believed that congenital

It the 1980s, it was generally believed that congenital abnormalities affected about 7% of a large nation�s children.  Some people believe that the increase in number of chemicals in the environment has led to an increase in the incidence of abnormalities.  A recent study examined 351 randomly selected children and found that 50 of them showed signs of abnormality.  Is this strong evidence that the risk has increased? (We consider a P-value of around 5% to represent reasonable evidence.) Complete parts �a� through �f�.


Write appropriate hypotheses. Let p be the proport



Answer

General guidance

Concepts and reason
Statistical hypotheses testing:

Hypotheses testing is used to make inferences about the population based on the sample data. The hypotheses test consists of null hypothesis and alternative hypothesis.

Null hypothesis:

The null hypothesis states that there is no difference in the test, which is denoted by . Moreover, the sign of null hypothesis is equal, greater than or equal and less than or equal.

Alternative hypothesis:

The hypothesis that differs from the is called alternative hypothesis. This signifies that there is a significant difference in the test. The sign of alternative hypothesis is less than, greater than , or not equal .

Proportion: The ratio of the number of favorable outcomes and total number of possible outcomes is in the sample called as the proportion.

One sample z test:

The one sample z test is used to check the means of the two groups are compared. It is also known as Independent samples test. Example - “Does the population mean body temperature of men and women differ in a particular city”.

Z-statistic for proportion: The standardized z- statistic is defined as the ratio of the ‘distance of the observed statistic from the proportion of the null distribution’ and the ‘standard deviation of the null distribution’.

P-value: The probability of getting the value of the statistic that is as extreme as the observed statistic when the null hypothesis is true is called as P-value. Therefore, it assumes “null hypothesis is true”.

Fundamentals

The formula for sample proportion is,

Sample proportion(ə) =

Here,

x. number of success
nisample size

The test statistic for one proportion is,

z=
P-P.
Po(1-po)

Here,

P:Sample proportion
Po : Population proportion
n:Sample size

Conditions to perform the one-sample z-test:

1.The population follows normal distribution.

2.The sample size of population is larger than 30. That is, n> 30.

Assumptions of z test for a proportion:

1.The sample must be reasonably a random

2.The sample must be less than 10% of population.

3.The sampling method used is simple random sampling.

4.Each sample point having two outcomes. a success and failure.

5.The sample includes at least 10 successes and 10 failures.

6.The population size is at least 20 times larger than the sample size.

7.Observations should be independent.

The formula for the p-value for left-sided test is, p-value = P(Z <z)

Procedure for finding the probability is listed below:

1.From the table of standard normal distribution, locate the z-value

2.Move left until the first column is reached.

3.Move upward until the top row is reached.

4.Locate the probability value, by the intersection of the row and column values gives the area to the left of z.

Rejection rule for p-value method:

If p-value sa(=0.05), then reject the null hypothesis.

Z Test for one sample proportion.

Excel add-in (PHStat) procedure:

1.In EXCEL, Select Add-Ins > PHStat > One-Sample Tests > Z Test for the proportion.

2.Enter ---- under Null hypothesis.

3.Enter ---- under Level of significance.

4.Enter ---- under Number of Items Interest.

5.Enter ---- under Sample size.

6.In Test Options, choose Two (or Upper or Lower) Tail Test.

7.In Output Options, enter a Title and click OK.

The conditions for the normal approximations are as follows:

1.013 du

2.01 (d-1)

Where,

p:probability of success
n:Sample size

Step-by-step

Step 1 of 6

(a)

The null and alternative hypothesis are stated below:

Null hypothesis:

H,:Po =0.07

Alternative hypothesis:

H.: P. >0.07

Part a

Thus, the null and the alternative hypotheses are,

H,:Po =0.07 vs H.: P. >0.07


The null Hypothesis states that there is evidence that 7% of the nation’s children have genetic abnormalities. The alternative hypothesis states that there is evidence that more than 7% of the nation’s children have genetic abnormalities.

Step 2 of 6

(b)

The satisfied assumptions are identified as shown below:

From the given information, generally believed that congenital abnormalities affected 7% of nation’s children. In a recent study of 351 samples 50 are showed signs of abnormality.

That is, n=351andp=0.07.

Checking condition 1:

np = 351(0.07)
= 24.57
> 10

Thus, condition 1 is satisfied.

Checking condition 2:

n(1- p) = 351(1-0.07)
= 351x0.93
= 326.43
> 10

Thus, condition 2 is satisfied.

The sample should be collected randomly and observations would be independent. The sample size must be less than 10% of population size. However, each point of the test consists of only two out comes that are Failure or success and sample needs at least 10 successes and 10 failures. The population size is at least 20 times larger than the sample size.

Part b

The necessary assumptions are,

Less than 10% of the population was sampled.

The sample is random.

The data are independent.


The assumptions are obtained by using the one sample z-test for proportion. Hence, the sample should be less than 10% of the population and the sample should be random. The data should be normally distributed.

Step 3 of 6

(c)

The p-value is obtained as shown below:

From the given information, generally believed that congenital abnormalities affected 7% of nation’s children. In a recent study of 351 samples 50 are showed signs of abnormality.

That is, n=351x=50p=0.07 and a=0.05

Instructions to obtain the p-value.

1.In EXCEL, Select Add-Ins > PHStat > One-Sample Tests > Z Test for the proportion.

2.Enter 0.07under Null hypothesis.

3.Enter 0.05 under Level of significance.

4.Enter 50 under Number of Items Interest.

5.Enter 351 under Sample size.

6.In Test Options, choose Upper Tail Test.

7.In Output Options, enter a Title and click OK.

Follow the above instructions to obtain the following out

Z Test of Hypothesis for the Proportion
Data
Null Hypothesis AS
Level of significance
Number of Items of Interest
Sample Size

From the output, the p-value is 0.0000.

Part c

Thus, the P-value is 0.0000.


The P-value is obtained by substituting the sample size as 35, number of success as 50, level of significance as 0.05 and population proportion as 0.07 in Excel-PHstat.

Step 4 of 6

(d)

The P-value in this context is explained below:

The null Hypothesis states that 7% of the nation’s children have genetic abnormalities. The alternative hypothesis states that more than 7% of the nation’s children have genetic abnormalities.

In hypothesis problems null hypothesis will be tested using the given information and the P-value conclusion will also be based on null hypothesis

Hence, the p-value is the chance of observing 7% of children with genetic abnormalities.

Part d

Thus, the p-value is the chance of observing 7% of children with genetic abnormalities.


The P-value is based on the null hypothesis.

Step 5 of 6

(e)

The conclusion is stated below:

The p-value is 0.0000 and the level of significance is 0.05.

The p-value is less than level of significance.

That is, p-value (=0.00)<a(=0.05)

By the rejection rule, reject the null hypothesis.

Therefore, it can be concluded that there is sufficient evidence that more than 7% of the nation’s children have genetic abnormalities.

Part e

Thus, Reject the.There is sufficient evidence that more than 7% of the nation’s children have genetic abnormalities.


The conclusion is stated based on the p-value and the level of significance using the rejection rule. If the p-value is less than or equal to the level of significance, then the null hypothesis is rejected otherwise not. In this situation the p-value is less than the level of significance and thus the null hypothesis is rejected.

Step 6 of 6

(f)

The environmental chemicals cause congenital abnormalities or not is obtained below:

There is no information about environmental chemicals causing congenital abnormalities and the null Hypothesis states that 7% of the nation’s children have genetic abnormalities. The alternative hypothesis states that more than 7% of the nation’s children have genetic abnormalities.

Hence, it is unknown if environmental chemicals cause genetic abnormalities, because the hypothesis test does not indicate the cause of any changes.

Part f

Thus, it is unknown if environmental chemicals cause genetic abnormalities, because the hypothesis test does not indicate the cause of any changes.


It is unknown if environmental chemicals cause genetic abnormalities, because the hypothesis test does not indicate the cause of any changes.

Answer

Part a

Thus, the null and the alternative hypotheses are,

H,:Po =0.07 vs H.: P. >0.07

Part b

The necessary assumptions are,

Less than 10% of the population was sampled.

The sample is random.

The data are independent.

Part c

Thus, the P-value is 0.0000.

Part d

Thus, the p-value is the chance of observing 7% of children with genetic abnormalities.

Part e

Thus, Reject the.There is sufficient evidence that more than 7% of the nation’s children have genetic abnormalities.

Part f

Thus, it is unknown if environmental chemicals cause genetic abnormalities, because the hypothesis test does not indicate the cause of any changes.

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